Genetics 301 final Exam
Dr. F. Nargang
December 10, 2001
Name:______________________________
I.D.#:_______________________________
There are 19 questions and a total of 67 possible marks on this exam. You must do each question. Good luck!
The exam booklet should contain nine pages. The exam will constitute 40% of your final grade. TIME LIMIT
IS TWO HOURS.
Use diagrams whenever you wish. Don’t waste time by writing more than is needed to answer a question
correctly.
Question 1. From the assigned paper (Yamaoka et al. J. Biol. Chem. 155, 301 (2000)).
a) What method did they use to place mitochondria/mtDNA from one species into the cells of another?
(1 mark)
ρ0 cell of mice were fused with platelets of rats using PEG.
b) Why did they need to do the control experiment of repopulating their po cells with M.m. domesticus
(Mus musculus domesticus) mitochondria? (3 marks)
1. to show that the ρ0 cells were not simply mutants incapable of replicating any kind of mtDNA
2. to eliminate the possibility that the cybrids obtained in later experiments were not simply
revertant mouse cells that had indetectable amounts of their own mtDNA at the start of the
experiment but later recover and become ρ+ cells.
3. that the drug treatment for generating ρ0 cells did not also induce nuclear mutations affect mt
proteins necessary for oxphos (or respiration).
c) In the discussion they note that protein vs. nucleic acid interactions are less stringently controlled than
protein vs. protein interactions. What do they mean and what evidence supports this? (3 marks)
The mouse proteins needed for replication, transcription, and translation of the rat mtDNA and
the transcripts produced from it seemed to work fine since the rat mtDNA was replicated,
transcribed and mt proteins were translated. All these require protein/nucleic acid interactions.
However, the proteins produced from the rat mtDNA genes are not capable of correct interaction
with the nucelar encoded mouse proteins for the oxphos complexes since the activity of the
complexes and oxygen comsumption are low. The latter are protein-protein interactions.
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Question 2. A new species of unicellular organism has been discovered. Nothing is known about the mode of
mtDNA inheritance in the organism. It is known that it has some features similar to S. cerevisiae since it has
two mating types called b and β and it exists as a haploid, but can form a diploid. The diploid can grow and
divide normally or it can be induced to produce spores by placing it on specialized medium. Like many yeasts
and fungi it can use a variety of carbon sources for growth including glucose and glycerol. Growth on glycerol
requires the presence of oxygen. The diploid form makes buds only from the ends of the cell, not from the
middle region where the zygote formed.
Different strains of the organism are available isolated from the wild. The only detectable difference between
them is a polymorphism in their mtDNAs which distinguishes type 1 and type 2 mtDNA in the respective
strains.
a) You want to find out things about the inheritance and segregation of the mtDNA in this organism, so
you set up some crosses:
i) mating type b strain (type 1 mtDNA) x mating type β strain (type 2 mtDNA)
-if the diploids are allowed to undergo sporulation shortly after they are formed, the
mating type alleles segregate 50:50, but all progeny contain type 1 mtDNA.
ii) mating type b strain (type 2 mtDNA) x mating type β strain (type 1 mtDNA)
-if the diploids are allowed to undergo sporulation shortly after they are formed, the
mating type alleles segregate 50:50, but all progeny contain type 2 mtDNA.
What is the simplest interpretation of the data from these two crosses? (2 marks)
mtDNA is inherited only from the mating type b parent.
b) If the zygotes are allowed to undergo 25 mitotic divisions and the progeny are analyzed, you find that
there are a few cells with one type of mtDNA, but most contain both type 1 and type 2 mtDNA.
What do you conclude from these data? (2 marks)
Upon fusion the mtDNAs mix completely and randomly prior to first bud formation.
c) You now have isolated a mutant of the type 1 mtDNA. Strains containing the mutant mtDNA no
longer grow on glycerol. The results from Mendelian crosses with this mutant form are basically the
same as in a) above:
i) mating type b strain (type 1 mutant mtDNA) x mating type β strain (type 2 mtDNA)
-if the diploids are allowed to undergo sporulation shortly after they are formed, the
mating type alleles segregate 50:50, but all progeny contain type 1 mtDNA.
ii) mating type b strain (type 2 mtDNA) x mating type β strain (type 1 mutant mtDNA)
-if the diploids are allowed to undergo sporulation shortly after they are formed, the
mating type alleles segregate 50:50, but all progeny contain type 2 mtDNA.
However, if the zygotes are allowed to undergo 25 mitotic divisions and the progeny are analyzed, you
find that about 90% of the cells contain the mutant type 1 form of mtDNA.
What do you conclude from these data? Discuss how your conclusion explains the data. (3 marks)
The mutant type 1 mtDNA probalby had a replication or segregational advantage (similar to S.
cer. ρ- hypersuppressives) over type 2 mtDNA. Thus, during the 25 mitotic divisions, this type of
mtDNA gradually takes over the population. However in the sporulation experiment mtDNA is
inherited only from the type b parent. Perhaps the mtDNA from the β strain is actively degraded
OR the spores are produced before the mutant form could take over.
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Question 3. Explain why strains with the cytoplasmically inherited ability to suppress pet494 mutations contain
a mixture of two different types of mtDNA. (4 marks)
The deleted form allow cox3 to be translated without the requirement for the pet494 protein (Thus cox3
function is supplied).
The wildtype form supplies the functions that are missing as a result of the deletion (Thus all other
functions except for cox3 are supplied).
A complete set of functional mt products is required because the suppressors were selected on glycerol (ie
oxphos needed for growth).
Question 4. Multiple choice questions. There is one best answer for each question. (6 marks)
Mitochondria are formed from/by:
a) de novo synthesis in yeast buds
b) growth and division
c) entirely products of mitochondrial genes in Reclinomonas americana
d) entirely products of nuclear genes in some primitive mitochondria-containing protists
e) metachlorions
Which of the following contains a mt-encoded gene product in humans:
a) Krebs cycle enzymes
b) Complex I
c) Complex II
d) biochemical pathway for pyrimidine biosynthesis
e) Glycolysis enzymes
ATP is formed with the aid of:
a) F0 without F1 attached
b) a turning shaft in the small mt ribosomal subunit
c) a proton gradient across the mitochondrial inner membrane
d) a proton gradient across the mitochondrial outer membrane
e) superglue
The closest living relatives to the ancestors of mitochondria are:
a) Archaebacteria
b) primitive eukaryotes without mitochondria
c) α-proteobacteria
d) yeast ρo mutants
e) the malaria parasite
A respirasome is:
a) a rudimentary mitochondrion in primitive protozoan eukaryotes
b) a complex of two or more respiratory complexes
c) another name for the original endosymbiont
d) a mitochondrial ribosome synthesizing proteins required for respiration
e) another name for yeast mitochondria in cells growing on glycerol
The immediate consequences of deleting the CSBs from human mtDNA would be:
a) a decrease in transcription
b) an increase in transcription
c) increased D-loop formation
d) a decrease in catabolite synthesis blockers
e) an increase in electron transport
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Question 5. Define the following in one or two sentences: (6 marks)
a) segregational petite
A nuclear mutation affecting a mt component such that all cells are petite on medium containing high
glycerol/low glucose. The mutation segregates in Mendelian fashion.
b) complex V
The mt ATP synthase complex
c) omega
A gene in the intron of the large mt rRNA in some strains of yeast. It recognizes a sequence in the
corresponding region of strains that don't contain omega and inserts its sequence ther by a gene
conversion mechanism.
d) Mtg1 protein
Mitochondrial genome transmission protein. Required to resolve recombination junctions in yeast
mtDNA.
e) cytoplast
A cell whose nucleus has been removed.
f) cristae
The structure that results from infoldings of the inner mt membrane
Question 6. Below is a table showing a list of gene products in Saccharomyces cerevisiae. Complete the Table
by putting a check-mark in any correct boxes for each gene product. (4 marks)
Gene product
Citrate synthase
Large mt
ribosomal RNA
Cob
Most mt
ribosomal proteins
mt tRNAs
PET494
mt DNA
polymerase
Tom6p
Encoded
by
nuclear
gene
√
Encoded by
mitochondrial
gene
√
Resides in
mitochondrial
matrix
√
√
√
√
√
Resides in
mitochondrial
outer
membrane
Resides in
mitochondrial
intermembrane
space
√
√
√
√
√
Resides in
mitochondrial
inner membrane
√
√
√
√
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Question 7. Explain the “threshold effect” with respect to human mtDNA disease. (3 marks)
Mitochondrial function (or translation) is okay up to a certain low limit of wildtype mtDNA molecules.
After that point if the level of wildtype molecules decreases any further (below the threshold) mt function
is almost completely eliminated.
Question 8. Many genes encoding mitochondrial proteins are found in the nucleus and not mtDNA. Is this a
favorable situation for the cell or organism? Why or why not? (2 marks)
Probably favorable because genes in the nucleus undergo a sexual phase that includes recombination.
This allows the generation of individuals without mutations. mtDNA has no sexual phase and therefore
may accumulate mutations.
(Also acceptable: mtDNA may accumulated mutations more easily than nuclear DNA because of the
environment of oxygene byproducts in the mitochondria.
Question 9.
a) Describe the experimental evidence that suggests mtDNA in ρ+ x ρ+ crosses is fixed or anchored in the
portion of the zygote from which it is derived (at least until the time of first bud formation). (4 marks)
In the genetic portion of the experiments looking at fusion of individuals cells into zygotes, it was
observed that first buds formed from the ends of cells carried only the mtDNA (based on markers in the
genes of the mtDNA) from the parent that formed that half of the zygote.
c) Would you expect that hypersuppressive ρ− mtDNA is anchored until after first bud formation in a ρ− x ρ+
cross? Why or why not? (3 marks)
No. Because in hypersuppressive ρ- x ρ+ mt crosses the ρ- takes over in about 95% of the progeny. If the
ρ- mtDNA was efficiently anchored then first buds formed from the ρ+ side might form. These would go
on to give a significant portion of the population that were ρ+ progeny.
Question 10. Which type of mutant would you expect to lack more mitochondrial gene products, mit mutants or
syn mutants? Explain your answer. (2 marks)
mit mutants affect single mt genes involved in oxphos complexes. Syn mutants affect mt translation
therefore syn mutant would be missing more products.
Question 11. For S. cerevisiae we say that after 20-25 mitotic divisions in the mitochondrial cross:
ρ+ (oliR capS) x ρ+ (oliS capR), that all the progeny contain only one type of mtDNA. We would also say that
some of the cells derived from this mitochondrial cross are ρ+ (oliR capR). Are these statements in conflict?
Explain why or why not. (2 marks)
Not in conflict. Recombination of mtDNA molecules in the neck of the zygote could create recombinant
molecules containing both resistance markers. After 20-25 generations some cells would contain only this
type of mtDNA.
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Question 12. Describe two observations/findings that support the existence of mtDNA nucleoids in S.
cerevisiae. (2 marks)
The number of generations required to give homoplasmic cells in crosses involving two types of mtDNA
is much fewer than would be required based on the number of individual mtDNA molecules in the zygote.
The number of mtDNA molecules observed by microscopy (chondriolites) is lower than the number of
mtDNA molecules that are calculated to exist in a cell.
Question 13. In experiments monitoring mitochondrial fusion with GFP, why was it important to have the
CS-GFP fusion protein under the control of a galactose promoter? (1 mark)
So that expression of the fusion protein can be shut off just prior to zygote formation. This insures that
no GFP is imported into the mitos of the non GFP fusion parent mitochondria following the fusion event.
Question 14. Why do human ρo cells require a pyrimidine? (1 mark)
Because dihydroorate dehydrogenase (an enzyme needed for pyrimidine biosynthesis) needs a high mt
membrane potential in order to function. ρ0 cells have a very low mt membrane potential.
Question 15. Why is it unusual/unexpected that a mtRNA polymerase would resemble that of T7
bacteriophage. (1 mark)
Because mitos were derived from a bacterium and would be expected to have enzymes/proteins related to
bacteria.
Question 16. In Saccharomyces cerevisiae mtDNA, would it be possible to have a mtDNA mutation in an
intron sequence (not at the exon/intron junction) that affected expression of the exons of that gene? Why or why
not? (2 marks)
Yes. Because some S.c. mtDNA genes contain introns encoding maturase function. This function is
needed to remove the intron. Thus if an intron mutation removed the function of the maturase, the exons
would not be properly expressed.
Question 17. If all the mt genes in a certain species of eukaryote were to eventually migrate to the nucleus so
that the cell contained no mtDNA, would you expect that mitochondria would eventually be lost from this
species? Why or why not? (2 marks)
No. The cell would still need mitos to carry on oxphos and other metabolic processes that occur in the
mito.
Question 18. Why did the experiments using GFP and TMR-CH2Cl (mitotracker) labeled mitochondria prove
that mitos mixed their contents, while the experiments using antibody to citrate synthase only suggested it? (2
marks)
The double label experiment showed complete colocalization of the colors showing that they must be
fused and completley mixed. With the single label citrate synthase experiment, it is possible that whole
mitos or portions of them were moving to the other half of the zygote without fusion and mixing.
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Question 19.
a) For mammals, give two models/hypotheses that attempt to account for the appearance of homoplasmic
offspring form heteroplasmic mothers. For each, speculate if it would be possible or virtually impossible for a
mother with a 50/50 ratio of two mtDNA types to give rise to a homoplasmic offspring. (3 marks)
1. Very small bottleneck of mtDNA molecules in some stage of development that would give rise to egg
cells that were homoplasmic.
-possible to get a homoplasmic individual from a 50/50 mother.
2. A bottleneck with a larger number of molecules (~200). Random segregation of molecules in
developing germline can give rise to homoplasmic offspring from heteroplasmic mother.
-however if mother was 50/50, it seems far less likely to get a homoplasmic offspring.
b) If mtDNAs in the female germline existed in large nucleoids that tended to contain only one of the two types
of mtDNA molecule in a heteroplasmic individual, would this change the likelihood of obtaining a
homoplasmic individual from a 50/50 heteroplasmic mother in either model? Explain. (3 marks)
In both cases this would improve the chances since the total number of mtDNA molecules segregating
through the bottleneck could be large in number but smaller effectively in segregating units.
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