Download Problem set 8 answers

Problem set 8 answers
1. Integrins are alpha/beta heterodimeric receptors that bind to molecules
in the extracellular matrix. A primary role of integrins is in adhesion,
ensuring that cells attach to the matrix, and cell biologists have identified
several proteins that link the intracellular tail of the beta integrin to the
actin cytoskeletin. During the development of the Drosophila wing, cells
on the dorsal and ventral surfaces adhere to the matrix. Mutations in the
beta integrin gene lethal(1)myospheroid lead to recessive lethality, but
hypomorphic mutations lead to wing blisters where the portions of the
dorsal and ventral surfaces separate. From what you now know about
integrins, propose a model for how integrins function in wing
development.
Integrins are the receptors than mediate adhesion between the dorsal
and ventral surfaces of the wing.
You screen for mutations that lead to wing blisters using FLP/FRT
induced recombination. Describe which genetic tools you will need to
conduct the screen.
You will need an enhancer to drive FLP expression in the wing and
chromosomes with FRT near the centromeres. A cell autonomous marker
will also help define which cells are mutant.
2. To study DNA repair mechanisms, geneticists isolated yeast mutants
that were sensitive to various types of radiation; for example, mutants that
were more sensitive to UV light. Ten haploid mutants were isolated.
a. How would you do the complementation tests for these mutants?
A complementation test would be done by crossing the haploid strains
and scoring the phenotype in the diploids.
b. Based on the complementation chart below, how many genes are
defined by these mutations? + indicates complementation, - failure to
complement.
1
2
3
4
5
6
1
-
2
-
3
+
+
-
4
+
-
5
+
+
+
-
6
+
+
+
+
+
-
Three genes: 1, 2, and 4 are in one gene, 3 and 5 in another, and 6 in a
third gene.
c. The gene defined by mutation 1 was cloned. Overexpression of the wildtype gene reduces the UV sensitivity of either mutant 3 or mutant 6.
Describe two models to explain the genetic interactions.
The gene could be in the same pathway. In this case the gene defined by
mutation 1 would be downstream of the genes defined by mutations 3
and 6 since its overexpression can bypass the UV sensitivity of the other
mutations. Alternatively, the gene defined by mutation 1 could be in a
parallel pathway and its overexpression can bypass the requirement for
the pathway(s) defined by the mutations 3 and 6.
3. The sup-7 mutation of C. elegans is an amber suppressor tRNA that
inserts tryptophan at UAG amber codons. You isolate 20 mutations in the
unc-22 gene, which lead to a recessive phenotype causing worms to be
uncoordinated in their movement. Three of the mutations are suppressed
by the sup-7 mutation, but when you clone the gene and sequence the
mutant alleles, eight mutations resulted in amber stop codons. Why do
you suppose that the sup-7 suppressor didn't suppress some of these
mutations?
Placing a tryptophan in those positions did not yield a functional
protein.
4. Animals homozygous for the sup-7 mutation die at 15o C. In 1981, Bob
Waterston mutagenized sup-7/sup-7 hermaphodites, grew them at 25o C,
and shifted the F2s to 15o C. Several strains were isolated that now could
grow at 15o C. The suppressors of the sup-7 lethality were inseparable by
recombination from the sup-7 mutation. Strains carrying sup-7 and the
tightly linked suppressor (suppressor of a suppressor) no longer
suppressed amber alleles of genes that were suppressed by the original
sup-7. Animals homozygous for the sup-7 suppressor chromosome appear
wild type. From what you know about nonsense suppressors explain
these results. (Note: these suppressors were too frequently isolated to be
explained by the reverting the mutant nucleotide in sup-7 back to the wildtype nucleotide, which should be a rare event.)
The suppressor eliminates the function of the sup-7 gene, and thus it is
no longer an amber suppressor. The animals homozygous for this
mutation are wild type because there are other tRNATRP genes that can
function in its absence.
5. You are working with a pure breeding stock of white-eyed flies and
notice a spontaneous red-eyed male. You generate a pure breeding stock
of the revertant in additional crosses. Design a crossing scheme to
distinguish between two possible explanations for the mutation leading to
the red-eyed male: an intragenic suppressor that restores the function of
white gene and an autosomal extragenic dominant suppressor. Assume
that if the suppressor is in a second gene that the recessive phenotype of
the suppressor in the absence of the original white mutation leads to red
eyes.
We are trying to distinguish between a suppressor mutation that is
intragenic from one that is extragenic. There are several ways to do this.
One simple way is to cross the suppressed red-eyed males to white-eyed
females. If the suppressor mutation is intragenic, all the F1 males will
have a mutant white allele from their mother and have white eyes. If the
suppressor mutation is on an autosome, all the F1 males will have a
mutant white allele from their mother, but will also inherit the
dominant suppressor from their father and have red eyes.
6. The C. elegans lin-14 gene controls the timing of development in C.
elegans. LIN-14 protein is high early in development and gradually
decreases as development proceeds. lin-14 is defined by both dominant
and recessive mutant alleles. Animals that are homozygous for recessive
alleles develop precociously (developmental events occur much earlier
than normal) because LIN-14 protein levels are lowered or eliminated,
similar to the levels seen later in development. Animals that contain
dominant alleles are retarded in development (developmental events occur
much earlier than normal) because LIN-14 levels are higher than they
should be late in development. You are given three lin-14 mutants.
Animals that are homozygous for loss-of-function alleles of the lin-4
gene are retarded in development. The lin-4 RNA is thought to bind to the
lin-14 mRNA and inhibit translation of the LIN-14 protein. What genetic
epistasis result would be consistent with lin-4 functioning as a negative
regulator of lin-14.
The lin-4(lf); lin-14(lf) double mutant should develop precociously.
7. Kin1p and Kin2p are protein kinases that act sequentially in a "kinase
cascade.” Kin1p places a phosphate group on Kin2p, and this increases
the activity of Kin2p so that it may, in turn, place a phosphate group on its
target. Two kin1 mutants are isolated. One suppressor of each kin1 mutant
is then isolated. Both map to the KIN2 gene. One kin2sup mutant shows
allele specificity, only suppressing the kin1 allele that it was originally
selected to suppress.
a) Propose a model to account for these observations. What is the nature
of the original kin1 mutations and how can the kin2 mutants be allelespecific suppressors of these mutants?
The original mutation disrupts the ability of the Kin1 and Kin2 proteins
to interact, and the suppressor restores their ability to interact.
b) The other kin1 suppressor suppresses all kin1 alleles. How do you think
that this suppressor acts?
The mutant Kin-2 protein is active in the absence of functional Kin-1
protein.
8. The supply of nitrogen regulates yeast genes affecting nitrogen
catabolism (Remember the prion lectures). The Ure2 protein of S. cerevisiae
is a negative regulator of nitrogen catabolism that inhibits the Gln3
protein, a positive transcriptional regulator of genes involved in nitrogen
assimilation. For example, the Gln3 protein activates the transcription of
DAL5, which encodes allantoate permease, a cell surface molecule that
transports this alternate nitrogen source under poor nitrogen conditions.
In the presence of a good nitrogen source like ammonia, Ure2 protein is
active and inhibits Gln3 activity, whereas in a poor nitrogen source such as
allantoate, Ure2 is not active and Gln3 is active. Overproduction of the
Mks1 protein allows DAL5 expression on ammonia, and lack of Mks1
protein prevents DAL5 expression on allantoate. In addition, an mks1 ure2
double mutant (both are null mutations) expresses DAL5 on either
ammonia or allantoate.
Describe a linear genetic pathway for the function of GLN3, MKS1, URE2
and DAL5, and describe which genes are active or inactive in ammonia
and in allantoate.
9. The yeast SUP35 and SUP45 genes encode proteins that are involved in
translational termination at stop codons. Mutations in these genes result in
low levels of readthrough at stop codons, and were isolated as suppressor
mutations. Do you think that the sup35 and sup45 mutations are gene
specific or nonspecific? allele specific or nonspecific? Explain your
reasoning.
Gene nonspecific because sup35 and sup45 should suppress nonsense
mutations in other genes.
Allele specific because sup35 and sup45 should not suppress mutations
other than nonsense mutations, like missense and frameshift alleles.
10. Sexual development in C. elegans is controlled by the X:autosome ratio.
In XX animals the ratio is 1.0, resulting in hermaphrodite development; in
X0 animals the ratio is 0.5, resulting in male development. Amorphic or
null mutations in the genes tra-1 and tra-2 result in the transformation of
XX animals into phenotypic males. Gain-of-function mutations in tra-1
result in the opposite transformation: XO animals are transformed into
phenotypic hermaphrodites. Double mutant combinations between tra2(lf) and tra-1(gf) mutations have placed tra-1 downstream of tra-2.
Recessive mutations in the fem-1 gene result in the transformation of
XO animals into phenotypic females. Remember that C. elegans
hermaphrodites are females that produce sperm for a short term. The fact
that fem-1 XO animals are transformed into females and not
hermaphrodites has you puzzled. You think that perhaps the mutations
are hypomorphic mutations and null alleles would result in
transformation of XO animals into hermaphrodites.
a. You perform a genetic test to determine whether the mutations are
hypomorphic or amorphic. Describe the genetic experiments and the
results that would suggest that the fem-1 mutations are amorphic and not
hypomorphic.
Compare fem-1/fem-1 XO animals to fem-1/Df XO animals. The
phenotypes should be the same. In this case, both types of XO animals
will be transformed into females.
b. fem-1 acts between the two tra genes in the sex determination pathway.
Propose a genetic pathway for these three genes and define the activity
states (ON/OFF) of the genes in XX and XO animals. Describe the fem-1;
tra double mutant combinations that you would use to demonstrate this
pathway, and describe the karyotypes and the phenotypes of the double
mutants that support the proposed pathway.
fem-1; tra-2(lf) XO animals will be females. fem-1; tra-1(lf) XX animals
will be males.
11. A deletion in the p53 gene causes a dominantly inherited form of
cancer. Normal cells of an individual with cancer contain both the normal
and deleted forms of the p53 gene, but the cancer cells contain only the
deleted version. Explain these observations.
p53 is a tumor suppressor gene. While the cancer trait is inherited in a
dominant fashion, all p53 function must be lost for cells to become
cancerous. The loss of heterozygosity is caused either by mitotic
recombination, loss of the normal chromosome or a spontaneous
mutation in the wild-type p53 gene.
12. Utpal Banerjee's lab isolated a dominant mutation in Sos (SosJC2) on
chromosome 2 as a suppressor of a hypomorphic sev allele known as sevE4.
a. The investigators conducted a series of dosage experiments to determine
the nature of their dominant suppressor. The data are below.
Genotype
sevE4 / sevE4 ; Sos+ / Sos+
sevE4 / sevE4 ; SosJC2 / SosJC2
sevE4 / sevE4 ; SosJC2 / Sos+
sevE4 / sevE4 ; SosJC2 / Df
Level of suppression
0%
36.0%
16.4%
4.0%
What type of mutant allele is the SosJC2 allele? Be as specific as possible,
and explain your reasoning.
Hypermorphic, because the suppression of sev
enhanced by adding a wild-type allele (sev
E4
E4
/ sev
/ sev
E4
E4
; Sos
; Sos
JC2
JC2
/ Df is
+
/ Sos ).
b. Mosaic analysis of loss-of-function alleles of Sos isn't informative
because these alleles result in a cell lethal phenotype (there are no mutant
clones of Sos cells because the cells die). The SosJC2 allele, however, is not
cell lethal. Animals of the genotype w sevE4 / w sevE4 ; SosJC2 P[w+] / Sos+
were irradiated with X-rays, and mosaic ommatidia scored. Below are the
results.
Cell
R1
R2
R3
R4
R5
R6
R7
R8
wild type for w
57
39
42
46
46
57
69
41
mutant for w
12
30
27
23
23
12
0
28
These are the results for 69 mosaic ommatidia scored that had R7. Each
number represents the number of cells that are mutant or wild type for
white (w). In the 69 ommatidia analyzed, for example, 57 of the R1 cells
were wild type for white and 12 were mutant.
Describe a strategy that could have been used to insert the P[w+] transgene
into chromosome 2? Be specific and use crosses if necessary.
Inject white embryos (w/w or w/Y) with w+ cloned between P element
inverted repeats and with a plasmid that contains the trasposase gene.
The injected animals are crossed to white mutants, and the F1 progeny
screened for animals with red eyes. These should have w+ inserted into
the chromosomes at random positions. One of the w+ insertions that
mapped to chromosome 2 was used here.
What is the purpose of the P[w+] transgene?
The P[w+] transgene is used to mark the mutant chromosome so that
cells that are homozygous for Sos+ can be identified by their white
phenotype.
What is (are) the genotype(s) of the w+ clones.
E4
w sev
E4
/ w sev
; Sos
JC2
+
+
P[w ] / Sos
E4
or w sev
E4
/ w sev
; Sos
JC2
+
JC2
P[w ] / Sos
+
P[w ]
What is (are) the genotype(s) of the w clones.
E4
w sev
E4
/ w sev
+
+
; Sos / Sos
Do you think that Sos is acting cell autonomously or nonautonomously?
Explain your reasoning.
Cell autonomously. None of the mosaic ommatidia with R7 were
JC2
homozygous for wild-type Sos, indicating that the Sos
mutant needs
to be present in R7 for R7 to be generated.
13. In FLP/FRT mosaic screens for mutants with white eye clones that had
expanded and overgrown at the expense of red eye tissue to generate
unusually large eyes, several tumor suppressor genes were identified on
chromosome 3. As we discussed in class, these genes were interesting
because they normally inhibit cell division and promote apoptosis. In these
same screens the investigators also identified a second class of mutants. In
these mutant animals, the eyes were unusually large because there were
too many cell divisions just like the mutants that we discussed. The
difference is that the red eye tissue is also overgrown in this second class of
mutants. Expression of the Unpaired secreted ligand is increased in the
mutant cells. Unpaired is a signal in a conserved pathway that regulates
the transcription factor STAT. Loss of one copy of the STAT gene can
suppress the overproliferation phenotype caused by the loss of this second
class of tumor suppressor genes.
a. Describe four genetic elements/mutations present on the chromosomes
(not including the induced mutations) that are used to generate and detect
the mosaic eyes and define which cells are homozygous for the
mutagenized chromosome in these screens.
white mutation
FRT sites
FLP driven by the eyeless promoter
P[white+] transgene
The white cells are homozygous for the mutagenized chromosome.
b. In the original screen, what are the genotypes of the white cells and the
red cells of the mosaic females that have the second type of tumor
suppressor gene mutation?
The white cells are eyeless-FLP w/w; FRT TS/FRT TS where TS is the
tumor suppressor mutation.
The red cells are either eyeless-FLP w/w; FRT TS/FRT P[w+] or eyelessFLP w/w; FRT P[w+]/FRT P[w+]
c. Describe a model that can explain the overgrowth phenotype of this
second class of tumor suppressor that incorporates the tumor suppressor,
Unpaired and STAT, defining in which cells the genes act and how they
might regulate one another. No molecular details are necessary.
TS inhibits Unpaired expression, and Unpaired activates STAT. TS and
Unpaired act in the same cells to send a signal to surrounding cells.
STAT acts in the responding cells.