Name_________KEY_____________________
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MCB140: Second Midterm
Fall, 2007
Before you start, print your name and student identification number (S.I.D)
at the top of each page. There are ten pages including this page.
You will have 120 minutes for the 150-point exam. The value of each question
is given at the beginning of the question.
Place your answer on the front of the page. Only answers in that space will
be graded.
You may write in pencil; however, to preserve your rights to a regrade, you
must write your answers in pen.
Good luck!
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This section is for grading. Do not write here.
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EXAM SCORE: __________
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Name_________KEY_____________________
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1. (10 points) You are given ten haploid petite yeast strains and told that they are either
nuclear petite mutants or - mtDNA mutants. You cross two of the strains and find that all
the diploids are petite. Give all plausible explanations for this result.
You could have crossed two nuclear petites that are mutant for the same gene. Or you
could have crossed two - petites.
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Name_________KEY_____________________
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2. (35 points) The left and right ASE chemosensory neurons express different cell surface
receptors that detect particular chemicals in the environment. For example, the right ASE
(ASER) but not the left ASE (ASEL) expresses the receptor GCY-5. To conduct a screen for
mutants defective in this asymmetric control of GCY-5 receptor expression, Oliver Hobert
and his colleagues generated a transgene where GFP was under the control of the gcy-5
promoter. In these transgenic animals, GFP was expressed in only the ASER. The screen was
designed to identify genes that were normally involved in inhibiting gcy-5 expression in
ASEL Describe how the screen was conducted for zygotic mutants (not maternal effect
mutants). (10 points)
Wild-type hermaphrodites carrying the reporter transgene would have been
mutagenized and selfed for two generations. The F2 progeny would have been screened
for mutants that express GFP in both ASE neurons.
If you wanted to identify maternal-effect mutations, how would you modify your screen?
Explain your reasoning. (5 points)
You would have screened the F3 progeny of the mutagenized hermaphrodites. Maternal
effect mutants would not have been isolated in the F2 because their F1 parent would
have been heterozygous for the mutation. However, the F2 progeny can be homozygous
for the mutation, and since it is the genotype of the mother that makes a difference for
maternal effect genes, her F3 progeny will exhibit the mutant phenotype.
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Name_________KEY_____________________
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The screen identified mutations in a number of genes, including a recessive mutation in the
gene lys-6 and a dominant mutation in gene cog-1. In a related screen, recessive mutations in
cog-1 produced the opposite phenotype: neither ASE expressed GFP from the gcy-5
promoter. Genetic epistasis experiments indicated that lys-6 is a negative regulator of cog-1.
Describe the experiments and their outcome. (10 points)
Animals mutant for loss-of-function mutations in both lys-6 and cog-1 will not express
GFP in either ASE, the cog-1 mutant phenotype.
The lys-6 mutation is a deletion that removes a gene that encodes an RNA but not a protein.
The cog-1 gene encodes a transcription factor (transcription factors can activate or repress
transcription of specific genes). Describe the simplest molecular model for how lys-6 and
cog-1 regulate the gcy genes. Also speculate on where the cog-1 gain-of-function mutation is
in the gene and how it affects its regulation. (10 points)
The lys-6 gene encodes a miRNA that inhibits translation of the cog-1 mRNA by
binding to the cog-1 3’UTR. The gain-of-function mutation in cog-1 would be in the
3’UTR and would disrupt the interactions between the #3’ UTR and the lys-6 miRNA,
allowing the COG-1 protein to be translated in ASEL.
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Name_________KEY_____________________
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3. (65 points) The Boss ligand binds to and stimulates the Sevenless receptor tyrosine kinase.
In boss and sev mutants, the flies are missing R7 and can’t detect UV light. In an F1 screen
for Boss mutants that can detect UV light, the allele known as Sevenmaker was isolated. As
the name indicates the boss-/boss-; Sevenmaker/+ animals sometimes had one to several R7
cells in ommatidia. That’s why they can now see UV light. The Sevenmaker allele is a gainof function allele. How would you show that Sevenmaker is a gain-of-function allele?
Explain your logic. (5 points)
Look at the R7 phenotype for animals that are hemizygous (Df/+) for the Sevenmaker
locus. If Sevenmaker is a gain-of-function mutation, the hemizygous animals should not
have extra R7 cells.
If you compare Sevenmaker/+ to Sevenmaker/Df animals, how would the animals compare if:
Sevenmaker is a hypermorphic mutation? (5 points)
Sevenmaker/+ animals would have a more severe phenotype. The would have more
ommatidia with extra cells or ommatidia with more extra cells.
Sevenmaker is a neomorphic mutation? (5 points)
Sevenmaker/+ animals would be similar in phenotype to to Sevenmaker/Df animals
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Name_________KEY_____________________
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Sevenmaker is a gain-of-function allele of a gene called rolled. Most recessive mutations in
rolled result in lethality. One recessive allele of rolled results in animals that survive to
adulthood, and the ommatidia of these animals occasionally are missing R7. You suspect that
this allele is hypomorphic. How would you test this idea? Explain the result if the allele is
hypomorphic. (5 points)
Ask whether the rolled allele over a Df (or over a lethal allele) would have a more severe
phenotype. More of the rolled/Df animals would be missing R7 cells (or they would be
dead is another answer).
The Sevenmaker allele can produce one or even multiple R7 cells in the boss and sev
homozygous mutant background, but in a sina homozygous mutant background Sevenmaker
doesn’t generate R7 cells. In the absence of Sevenmaker, recessive mutations in sina also
lead to a R7 missing phenotype. Like sev and boss, sina acts autonomously in the R7 cell.
Assuming that the genes act in a regulatory pathway, describe the order and how the sev, sina
and rolled genes act. (10 points)
Sevenless positively regulates Rolled , which positively regulates Sina.
Do you think that an amorphic allele of rolled could be a dominant enhancer or suppressor of
the sevts allele we talked about? Why? (5 points)
It could be a dominant enhancer of sevts because haploinsufficiency of a downstream
component of the Sevenless signaling pathway could produce an R7 missing phenotype
in sevts animals grown at the permissive temperature (or could say at 22.3o).
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Name_________KEY_____________________
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You want to define what cell requires rolled for its function in specifying the R7 fate, but the
hypomorphic allele produces an R7 phenotype that is too impenetrant for the experiment and
the stronger alleles are cell lethal. You decide to use the Sevenmaker allele. How would you
conduct the experiment? Make sure that you describe all of the necessary tools, the genotype
of the animal that you are using for the mosaic analysis (you don’t need to explain how to
construct this strain-no crosses are necessary), and the expected outcome of the experiment.
Hint: boss (or sev) must be homozygous for this experiment to work. (30 points)
The genotype of the animals would be Peye-FLP sev w/sev w; FRT
Sevenmaker P[w+]/ FRT + (or they could be homozygous for boss).
Peye-FLP and FRT don’t need to be present if they induce recombination by
X rays. Need to say that recombination induced while eye developing.
You would look at mosaic ommatidia, and you would expect Sevenmaker/
Rolled to act autonomously because it acts between to Sev and Sina that act
autonomously. Thus if an R7 is present in a mosaic ommatidium, then it
must be red. All of the other photoreceptor cells can be white, and a red R7
cell can still be present. All of the other photoreceptor cells can be red, and
R7 could still be missing.
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Name_________KEY_____________________
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4. (30 points) I lectured about the experiment where in an attempt to isolate new alleles of the
coat color gene C, wild-type male mice with black fur were mutagenized with X rays and
mated to albino female mice that were homozygous for the recessive allele c. Many albino
mice were isolated in this screen. Explain how they arose. (5 points)
Mutations in the C gene induced in the germline of the mutagenized male failed to
complement the albino mutation.
Also, a few mice with grey or off-colored white coats were isolated. Explain these progeny.
(5 points)
The same as the white progeny except that the new mutation was a weak (hypomorphic,
partial loss-of-function) allele of C.
The most intriguing mouse was a variegated female, having patches of white and black fur.
As discussed, this mutant contained a reciprocal translocation, where a portion of the
autosome that contained the C gene was translocated onto the X chromosome. Explain why
this female was variegated and what happened in the cells that produce black and white fur.
(10 points)
The female was variegated because of X-inactivation. When the translocation
chromosome containing the piece of the autosome with the C gene is inactivated, then
those cells will be mutant for C and generate only white fur. If the nontranslocation X is
inactivated, the C+ allele on the translocation X chromsome is expressed and these cells
generate black fur.
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Name_________KEY_____________________
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You cross the variegated female to an albino male. What will be the phenotypes of the
progeny? (10 points)
The males will black if they inherit the translocation chromosomes and white if they
inherit the normal chromosomes.
The females will variegated if they inherit the translocation chromosomes and white if
they inherit the normal chromosomes
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Name_________KEY_____________________
SID_____________________
5. (10 points) You are given three fly stocks. Two are wild type (labeled A and B) and one is
homozygous for a nonsense mutation in the autosomal gene boss, causing the animals to lack
R7 cells. You are told that one of the wild-type stocks is a laboratory M strain, and the other
is a P strain from the wild. How will you determine which is which? Unfortunately, your
high temperature incubator is broken so you can’t use temperature-sensitive sterility.
Cross the A and B strains in both orientations, and then cross the F1 progeny to
the boss/boss mutants and the F2 are screened for mutants missing R7, caused by
the formation of a new allele of boss being generated in the germline of the
hybrid dysgenic F1 progeny. If A is the M strain, then when it is crossed as a
female, you should see R7 mutants in the F2. None of these mutants will be
produced when A is crossed as the male. If B is the M strain, then when it is
crossed as a female, you should see R7 mutants in the F2. None of these mutants
will be produced when B is crossed as the male.
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