1
FINAL EXAM, 2011

Due: Friday, Dec 9, 5pm (NO LATE FINALS WILL BE ACCEPTED). Turn in
finals to the Agard lab, room S414, in the box on Elena’s desk.

Please type out your answers (you can draw in any diagrams but please use pen).

Have your name, organism, and page number on each page (example: John
Smith, Phage, Page 1 of 2).

Start your answers for each organism on a new page, and staple each organism
section separately.

Please be explicit and detailed in your answers in order to receive full credit.

Use correct nomenclature for gene and gene products.

You must work alone without help from other people, but you can use any
inanimate reference material.

If you need a clarification on any of the questions contact:
Angela: Worms, Flies
Elena: Prokaryotes, S.cerevisiae
Kate: Pombe, Mice
The TAs are not allowed to discuss course content during the final exam.
2
GENETICS OF PROKARYOTES
A. As mentioned in class, RecA and LexA are major components of the SOS response
that allow E. coli cells to respond to DNA damage. A number of interesting alleles of
the genes encoding these factors have been identified and were instrumental in
elucidating their function. Describe how you would screen/select for the following
alleles:
1. recA- (loss of function)
2. recAtif (tif = temperature-induced filamentation, an allele of recA that is activated
at 42C but not at 37C)
3. lexA- (loss of function)
4. lexAind (lexA that is uninducible by treatments with DNA damaging agents.
You can assume that both wild-type E. coli and any molecular genetic tool you would
need is available to you (except access to the recA and lexA mutant alleles in
question, of course). For each allele, describe a secondary test you would perform
that would increase your confidence that you have the desired allele. Indicate
whether your approach is a screen or selection, and whether you think each resulting
allele is dominant or recessive.
B. Fill out the following chart indicating whether the SOS response is ON (+) or OFF (-)
in the various mutants at 42C, and under both non-inducing conditions (-UV) or
upon activation (+UV):
E. coli genotype
WT
treatment
-UV
+UV
recA-
recAtif
lexA-
lexAind
recAlexA-
recAtif
lexAind
3
FUNGAL GENETICS
Saccharomyces cerevisiae
You are doing a screen for mutants affecting kinetochore function. You hypothesize that
these mutants will be sensitive to the microtubule depolymerizing drug Nocodazole.
Therefore, you look for Nocodazole sensitivity. You isolate 100 mutants (you are very
hard working, and like round numbers).
For all parts of all questions you won’t have cloned or sequenced any of these genes.
You will just do genetics, without identifying the genes that are mutated for each of
the complementation groups.
You get 4 complementation groups,
Group A: 47
Group B: 36
Group C: 16
Group D: 1
1) When you cross a mutant from one of the complementation groups to wildtype, you
see something funny in your tetrads. You get the following:
Nocodazole sensitive = N Sen
Nocodazole resistant = N Res (This is the wildtype condition for this concentration)
DEAD means the spore never grew into a colony
Note that you are initially growing the tetrads up on normal plates (no drug) and
subsequently replica plating them to the plate with Nocodazole. The dead spores didn’t
grow into colonies on the first, normal plate.
Tetrad 1
N Sen
N Res
N Sen
N Res
Tetrad 2
DEAD
N Res
N Res
DEAD
Tetrad 3
N Res
N Sen
N Sen
N Res
Tetrad 4
N Res
DEAD
DEAD
N Res
Tetrad 5
DEAD
N Res
N Res
DEAD
Tetrad 6
N Res
N Sen
N Sen
N Res
You cross this to a new wildtype, and it does the same thing, so you suspect that there
could be a second mutation in your original mutant strain.
A) What is the nature of the second mutation?
B) Give the likely genotypes for the first two tetrads. For ease of discussion let us call the
mutation responsible for the nocodazole sensitivity kin1-1 and we will call the second
mutation mut1-1, with the two wild types being KIN1 and MUT1 respectively.
C) Does the segregation suggest anything else about the location of the genes being
examined
4
D) Which complementation group is this mutant likely to be in and why?
2) You had a dream in which two sumo wrestlers stopped battling to unite their strengths
to fight a common enemy. You interpret this as a prophesy that two of the proteins
affected in your mutants are functioning together as a dimer. Pick mutants from three of
the complementation groups above (not including the one that gives the weird genetics in
part 1) and explain EXACTLY how you might conduct an experiment to look at this.
This will be a genetic experiment, not a biochemical one (since you don’t know what
genes are affected). Moreover, this experiment will not prove that these two gene
products act in one complex, but will provide evidence consistent with them functioning
in a mutually dependent way. Include in your analysis exactly how all strains would be
generated. That is, if you mention a strain of a given genotype, state how you would
isolate that strain, in addition to explaining the experiment you would do.
3) Gene conversion was discovered in yeast. Yet, there was years of work preceding this
studying genetics in peas, fruitflies and mice. Why wasn’t gene conversion found in
these organisms? (Hint, it isn’t just because yeast geneticists are better)
5
Schizosaccharomyces pombe
This question concerns the mechanisms by which cells maintain their gene expression
states. Grewal and Klar (Cell, 1995) made an intriguing observation when they deleted a
region of a S. pombe mat2/3 locus replaced with the ura4 gene (see diagram above).
Deletion of the K region in this manner resulted in cells that had two phenotypes:
FOA+/Ura- or FOA-/Ura+. (In contrast, an insertion of a ura4+ gene into the mat2/3
region results in silencing of the ura4+ gene). These cells occasionally switched from
one phenotype to the other. Deletion of clr1+, clr2+, clr3+, clr4+, or swi6+ silencing
genes in these strains resulted in a non-silenced phenotype (FOA-/Ura+). A cross was
performed (shown above) between a FOA+/Ura- strain and a FOA-/Ura+ strain. (Note
that both parental strains harbored a mutation in the endogenous ura4+ gene – this is not
shown in the genotypes shown above). Meiotic progeny were dissected and allowed to
germinate on rich media (YEA) and then replica-plated to –His, -Ura, or FOA media. In
addition, colonies were replica-plated to nitrogen starvation medium and stained with
iodine to detect sporulation, indicated by PMA+ above. Please answer each question
below using 1 or 2 sentences.
6
a. Based on these data, what would you conclude about the inheritance of the
silenced and expressed states of the ura4+ gene that was used to replace the K
region?
b. Following the segregation of the ura4+ gene expression state and the –His
phenotype, describe the tetrad type (PD, NPD, T) of each of the ten tetrads shown
above.
c. What is the significance of your answer to part (b) concerning the location of the
his gene?
d. The authors of this work argued that the switch between the ura4+ “on” and “off”
states represented an “epigenetic switch,” namely one in which a heritable change
in cell phenotype occurred that did not involve a change in the DNA sequence.
Use a diagram to suggest a mechanism by which DNA sequence changes could in
theory mediate a reversible but heritable change in the expression of a gene (note:
there are several possible mechanisms one could imagine – you need only
describe one, but you will receive extra credit for more than one)?
e. Experimentally, how would you use modern techniques to test your bioregulatory
hypothesis (or hypotheses) for a genetic mechanism by which the K::ura4+ gene
switches between its two states?
f. ura4+ genes are normally silenced when inserted into the silenced region of the
pombe mating type locus in a manner that does not delete sequences around the
insertion site. Experimentally how would you obtain such strains if one cannot
select for Ura+ transformants using your knowledge of silencing in S. pombe?
g. Extra Credit: What does the iodine staining phenotype tell you about the
relationship between mating type switching and silencing?
7
GENETICS OFTHE NEMATODE
C. elegans
You are studying programmed cell death, a conserved physiological process. C. elegans
mutants with excess or reduced programmed cell death are known. Loss-of-function
mutations in either ded-3 or ded-4 result in the survival of cells that normally die
indicating that these genes are required for the killing process. By contrast, a gain of
function mutation in ded-9 prevents most, if not all, programmed cell deaths while loss of
function mutations in ded-9 cause embryonic lethality. Thus, ded-9 is a negative
regulator of programmed cell death. Genetic and biochemical analyses established the
following epistatic relationship:
ded-9 --| ded-4  ded-3  programmed cell death
In C. elegans, a number of cell divisions/fate choices occur post-embryonically. One
example is the life/death decision of a pair of hermaphrodite specific neurons (HSN),
which innervate the vulva and are required for egg-laying. HSN neurons are born but
undergo programmed cell death in males. Hermaphrodites in which HSN neurons are
inappropriately killed are egg-laying defective (egl). A mutagenesis screen looking for
egl phenotypes identified egl-5(n1084) mutant animals in a standard F1/F2 screen. 100%
of egl-5(n1084) progeny exhibit an egl phenotype. You initiate backcrossing egl5(n1084) hermaphrodites with wild type males and obtain the following results:
P0:
F1:
egl-5(n1084)/egl-5(n1084) hermaphrodite X wild type males
You recover both males and hermaphrodites; ~75% of hermaphrodite
progeny exhibit the egl phenotype
1. What do you conclude about the nature of (n1084) mutation?
Using the egl phenotype you map egl-5(n1084) to a small region on chromosome V,
however, you still haven’t nailed down its molecular identity just yet. There is a
previously reported C. elegans strain, nDf42, which carries a deletion (a deficiency)
corresponding to this very region of chromosome V. The region corresponding to nDf42
deletes several tens of predicted genes. You obtain the following results in
hermaphrodites:
Table I.
Genotype
Wild type (+/+)
egl-5(n1084)/egl-5(n1084)
egl-5(n1084)/ +
nDf42/+
egl-5(n1084)/ nDf42
% Egg laying defective
1
99
71
1
68
%HSN surviving
100
0
27
92
42
8
2) Comparing egl-5(n1084)/+ and egl-5(n1084)/nDf42, do these results support or
refute your conclusion from question 1 regarding the nature of the egl-5(n1084)
allele? What do the results of nDf42/+ allow you to rule out? To answer these
questions, think in terms of recessive, dominant, haploinsufficient, etc.
3) To understand the role of egl-5 in programmed cell death, you are told to
examine loss of function mutations in egl-5. Why is that?
To isolate dominant suppressors of egl-5(n1084), you mutagenize egl-5(n1084)/egl5(n1084) hermaphrodites and screen through the resulting F1 progeny for rare animals
that display normal egg-laying and retain their HSN neurons. You identify a suppressor
(n3082) and determine that this suppressor is very tightly linked to the egl-5 locus. Now
that you have a suppressor strain, egl-5(n1084 n3082), you examine HSN survival as well
as survival of other cells that normally undergo apoptosis in hermaphrodites (same table
as above, now with extra data added):
Genotype
% Egg laying defective
Wild type (+/+)
1
egl-5(n1084)/(n1084
100
egl-5(n1084)/ +
71
nDf42/+
1
egl-5(n1084)/nDf42
68
egl-5(n1084 n3082)/(n1084 n3082) 0
egl-5(n1084 n3082)/ +
0
egl-5(n1084 n3082)/nDf42
1
%HSN surviving
100
0
27
92
42
97
98
100
extra cells ?
no
no
ND
no
ND
Yes
no
Yes
ND=not determined
4) What does the observation that the n3082 suppressor is tightly linked to egl5(n1084) suggest about the mutation?
5) From the data in the table above, what type of allele is n3082 likely to be (e.g. loss
of function, gain of function,…)? Given this, what do you conclude about the likely
role of egl-5 in programmed cell death and in HSN?
6) Now that you know the molecular identity of egl-5, suggest an easy experiment to
verify your conclusions regarding the role of this gene in programmed cell death
and HSN survival.
You identify the molecular identity of the egl-5 gene. It encodes a novel protein. To
determine the relationship of egl-5 and other components of cell death machinery, you
use P mec-7, a promoter that becomes activated only in a small number of cells, including
the six touch cell neurons of C. elegans. These touch cells are readily identifiable. These
9
cells are normally present in adult hermaphrodites. Animals that lack these cells are
viable but unresponsive to touch. Expression of wild type ded-3 or ded-4 under Pmec-7
promoter causes death of the touch cells (and ded-3, ded-4, and ded-9 are expressed
throughout the animal). You obtain the following data for various indicated transgenic
animals:
Table 2.
Transgene
% touch cells surviving
Pmec-7 (promoter alone)
100%
Pmec-7 egl-5(wild type)
8%
Pmec-7 egl-5; ded-9(gain of function)
100%
Pmec-7 egl-5; ded-4 (loss of function)
97%
Pmec-7 egl-5; ded-3 (loss of function)
98%
7) Based on these data draw a genetic pathway of how egl-5 may fit with the cell
death pathway (ded-9 --| ded-4  ded-3  programmed cell death).
It was previously shown that DED-9 normally binds DED-4. You now find that EGL-5
pulls down DED-9, however, any time that you use EGL-5 to pull down DED-9, you
don’t see DED-4 coming down with DED-9.
8) Describe in a few sentences a model that is consistent with the genetic and the
biochemical findings.
Extra credit: You determine that the n1084 mutation causes a point mutation within a
stretch of DNA downstream of the coding region of the egl-5 while n3082 causes a 5bp
deletion at the beginning of exon 2. The n3082 deletion is predicted to result in a frameshift leading to generation of a truncated protein composed of the first half of the wild
type protein and a 16 amino acid C-terminal extension in a different reading frame.
Moreover, you discover that FRAU-1, a transcription factor, binds to the DNA region
corresponding to the n1084 allele. FRAU-1 is expressed in the HSN neurons of
hermaphrodites but not males.
Come up with a model that is consistent with all of the data presented that could
explain how HSN neurons undergo programmed cell death in males but survive in
hermaphrodites. Include your conclusions/assumptions about the role of egl-5 in
programmed cell death, its activity in the HSN neurons, and the nature of the n1084
and n3082 alleles. Please include a diagram to illustrate your model.
10
FLY GENETICS
General Background:
You are interested in a curious sexual dimorphorism: the adult males of many higher
insects (this includes the diptera and Drosophila are dipteran – two winged) have one less
abdominal segment than the female. The larvae of both sexes have the same number of
segments but a specific segment is deleted during pupal development of males. The
segments of the adult are produced from nests of diploid cells that appear during
embryogenesis but remain quiescent (don’t divide) throughout embryonic and larval life
and then grow and divide rapidly at the beginning of pupal development. The cells in
these nests are called histoblasts. Each larval segment has four nests (a dorsal and a
ventral one, on both the left and right sides) that produce the corresponding parts of the
abdominal segment. Each nest has about 16 cells that have been arrested in G2, since
embryogenesis (i.e. during all larval stages). It is known that the male and female larvae
have the same number of nests. Since a specific segment fails to develop in males, some
sexually regulated modification oblates histoblasts in the segment that is deleted, or
perhaps curtails their development. You are interested in figuring out how this
regulation works.
In order to investigate the fate of the histoblasts, you need a way of positively marking
histoblasts to track them throughout pupal development. If you can mark them you can
follow the growth, development and possible elimination of the hisotoblasts from the
segment (A7) that is deleted in males to the histoblasts of the adjacent segment. By
tracking what happens in male and female pupae and in different mutants, you will be
able to figure out all sorts of things. But first, you have to mark the cells and for this
question that is the only thing you have to deal with.
Some specifics:
There is a gene called escargot (esg) that is a transcription factor expressed in the diploid
cells (i.e. not in larval cells which are polytene). The only cells in the abdomen of the
larvae that express esg are the histoblasts. You have an esg-GAL4 transgene that
expresses GAL4 in the histoblasts starting shortly after the histoblasts arise during
embryonic life. BUT esg expression (and the esg-GAL4 transgene) shuts off toward the
beginning of pupation and there is no residue of the earlier expression after the first two
of the pupal cell cycles; however, you want to track the cells through many cell cycles
and to follow their fate throughout pupal development. I would like you to describe an
approach that would allow you to mark the cells utilizing esg-Gal4.
For one third of the points – come up with a method to mark the histoblasts throughout
pupal development. For full points, 1) write a few sentences describing how your system
accomplishes this, 2) write out the genotype of the flies carrying your marker system that
you will observe (specify on which chromosome each element/gene you are using is
located), and 3) write a sentence or two describing each of the transgenes you are using
11
(their properties relevant to this problem - not all the details and not a discourse on Pelements).
For another third of the points - come up with one alternative way to mark the
histoblasts throughout pupal development (note that your first and alternative approaches
do not have to equally practical). For full points, 1) write a few sentences describing how
your system accomplishes this, 2) write out the genotype of the flies carrying your
marker system that you will observe (specify on which chromosome each element/gene
you are using is located), and 3) write a sentence or two describing each of the transgenes
you are using (their relevant properties - not their sequence and all the details). In
addition, of the ways you came up with, identify the better of the approaches and
indicate why it is better-suited to your goal.
Finally, to see how practical your marking system is, describe how it would be used to
address the following:
A gene called doublesex (dsx) located at 86E (on the right arm of the 3rd chromosome)
plays an important role in the regulatory cascade controlling sexual traits. The dsx
transcript is alternatively spliced in males and females. You have a mutant that
specifically inactivates the male isoform of dsx, dsxm. It behaves as a recessive mutation
that only affects males. The homozygous males are viable but exhibit some female traits
and they are sterile. Notably, the sterile males have an A7 abdominal segment (i.e. Dsxm
is required for the male specific segment elimination). You have a stock of this mutation
with a third chromosome balancer, TM6-Sb. Note that this stock has non-Sb flies that are
homozygous for dsxm, but the balancer is maintained in the stock because the only fertile
males are heterozygous. You want to follow the effect of dsxm on the fate of the
histoblasts in A7.
Some hopefully helpful notes:
When you are working out a crossing scheme, it is easiest to work backwards, describing
the genotype you want then defining the parents that will give the desired genotype and
grandparents that will give you the desired parents. Sometimes many generations are
required in elaborate strain constructions. Here, I want to limit the problem to the final
generation that produces the experimentally relevant progeny.
Note that the progeny that you will examine will be pupae not adult flies. Many of the
markers (e.g. Cy) used to follow genotype will not be scorable in the pupae. Fortunately,
GFP expression is visible, and, if you need them, there are a few markers that are
scorable in pupae (e.g. Tubby – a dominant 3rd chromosome marker that affects body
shape).
For the final third of the points, 1) Write out the genotype of the fly (actually a pupa) in
which you can observe the effect of dsxm on the histoblasts in A7, utilizing the BETTER
of the two systems you came up with. 2) Write out the immediate parents of this fly and
illustrate how you would cross them together to produce the fly in question. Draw out all
the possible progeny and how you would select for the proper genotype of the fly in
12
which you can observe the effect of dsxm on the histoblasts in A7. Feel free to come up
with various markers/Balancers, but please indicate their properties, especially if they are
of your own invention. Be sure to indicate the fraction of progeny exhibiting the desired
genotype and indicate if there are issues or tricks to identifying the desired progeny.
If you see issues that arise with your scheme (for example, you might find the crosses
cumbersome because the desired progeny class is rare, or the chromosomes are hard to
follow) identify what makes it difficult and propose adjustments that might make it
simple.
13
MOUSE GENETICS
Recent work has identified the recessive locus, hyde, underlying extreme aggression in
humans. The gene encodes a protein Hyde with no known functional domains. You are
working on social behaviors in mice and find by browsing on the ensembl site that the
gene is conserved across mammals. The mouse hyde locus looks like:
not to scale
Your in situ hybridization studies on the developing and adult brain show hyde to be
expressed only in the adult amygdala, a region known to be involved in aggression but
not essential for viability. Intrigued, you decide to knockout hyde in the mouse to assess
its role in behavior.
1) Design a gene targeting strategy to generate a constitutive null allele such that it also
labels Hyde expressing cells. Given that you have no idea what the two exons encode,
you will have to make a judgment call as to what construct is most likely to yield a null
allele. Please draw the construct and detail the targeting strategy and breeding needed to
obtain hyde/- mice.
After you have successfully obtained your hyde/- mice, you cross them to each other to
generate null animals (-/-) so that you can do some behavioral testing on them. However,
your crosses yield a smaller litter size and no null mice.
2a) Why this failure to obtain liveborn hyde null mice?
2b) How do you test your hypothesis?
2c) How do you reconcile these findings with the data from humans? Please provide at
least 2 different reasons.
You are still determined to test whether Hyde function in the mouse brain controls
aggressive type interactions.
3a) Please design a genetic strategy to do so. Assume you have access to the desired
promoters should your strategy call for one. Once again, draw the construct(s), detail the
targeting strategy, breeding, and any functional validation needed to generate the mice
you will use for your studies.
14
3b) Please detail the breeding required to generate your experimental and control mice
for the behavioral studies.