Download Practice exam (2010) key

PCB5065 Fall 2010
Exam 4
Total value = 70 points
Name _key____________________________________
Question 1
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Question 2
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Question 3
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Question 4
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Important – please keep your answers short; confine your answers to the space
provided; do not write on the back of any pages !
PCB5065 Fall 2010
Exam 4
page 2 of 5
Name ____________________________________________
1. (15 pt, 5 pt each part) The questions below pertain to the genetic transformation of
organelle genomes, which presents some special challenges.
a) Transgenes are introduced into organelles via bombardment with DNA-coated particles
and inserted into the organelle genomes via homologous recombination. Immediately after
this occurs, the bombarded cells are heteroplasmic for the transgene. What genetic behavior
of organelle genomes allows for the eventual selection of homoplasmic organelle genome
transformants? Explain your answer.
Somatic (mitotic) segregation
Through this process organelle genomes having different haplotypes sort into different cells.
Over multiple generations this leads to homoplasmic cells. In the absence of selection the
each of the initial haplotypes should be represented within the cell population. At this point
the even a recessive selectable marker, e.g. ribosome-based antibiotic resistance, could be
selected to eliminate the antibiotic sensitive haplotype and recover the resistant haplotype.
b) Because organelle genome transformation occurs by homologous recombination, mutant
correction by homologous gene replacement is possible. Explain how you would select for
yeast cells in which a mutant gene essential for mitochondrial respiration had been
successfully replaced with a wild-type gene.
wild type yeast cells can grow on glucose, via fermentation, or on glycerol, a carbon source
that can must be respired. Yeast mutants that cannot respire will grow on glucose but not on
glycerol, so cells could be tested for growth on glycerol or for large colony size on glucose +
glycerol media.
c) Because organelle genome transformation occurs by homologous recombination, mutant
correction by homologous gene replacement is possible. Explain how you would select for a
a chlamydomonas cell in which a mutant gene essential for chloroplast photosynthesis had
been successfully replaced with a wild-type gene?
wild-type chlamydomonas cells can grow in the presence of light without any supplied
carbon source in the media. (They fix CO2 via photosynthesis.) Chlamydomonas mutants
that cannot carry out photosynthesis will grow in the dark when provided acetate as a carbon
source, but will not grow in the presence of light and absence of acetate. So cells could be
tested for growth in the presence of light and absence of acetate.
or
Photosynthetic mutants of chlamydomonas are often light-sensitive due to damaging light
reactions that result from partially assembled photosynthetic complexes. So you could
select for cells that survive and grow in the presence of high light.
PCB5065 Fall 2010
Exam 4
page 3 of 5
Name ____________________________________________
2) (15 pt, 5 pt each part) In mice, the imprinted insulin growth factor 2 (Igf2) gene is
expressed from the paternally-inherited chromosome. The maternally contributed gene is
silent. The complete loss of Igf2 function is viable, the mutant mice are just much
smaller than wild-type mice.
a) If a mouse is heterozygous for a loss-of-function mutation at the Igf2 locus
(genotype Igf2 – / + ), will this mouse have a mutant or wild-type phenotype? Explain your
answer.
This depends upon whether the paternal or maternal parent contributed the mutant allele.
If the paternal parent was mutant, the moue in question will have a mutant phenotype.
If the maternal parent was mutant, the mouse will have a normal phenotype.
b) If an Igf2 – / + male mouse is mated with a wild-type (Igf2 +/+) female mouse, what are the
expected frequencies of Igf2 genotypes and resulting phenotypes in the offspring?
Explain your answer. Transmission is Mendelian but only the paternal allele is
expressed, so half are mutant and half wild-type
genotypes
frequencies
phenotypes
Igf2 – / +
0.5
small (mutant)
Igf2 +/+
0.5
normal (wild-type)
c) If an Igf2 +/+ male mouse is mated with an Igf2 – / + female mouse, what are the expected
frequencies of Igf2 genotypes and resulting phenotypes in the offspring? Explain your
answer. Transmission is Mendelian but only the paternal allele is expressed, so all are
wild-type, the mutant allele derived from the female is not expressed
genotypes
frequencies
phenotypes
Igf2 – / +
0.5
normal (wild-type)
Igf2 +/+
0.5
normal (wild-type)
PCB5065 Fall 2010
Exam 4
page 4 of 5
Name ____________________________________________
3. (20 pt.)
3a) (5 pt)The major gene classes that act in drosophila development are: homeotic
(segment identity) genes, gap genes, maternal effect genes, pair-rule genes and segment
polarity genes. In the table below, list these classes of genes in the order that they come
into play during the drosophila developmental program (first =1, last=5).
3b) (5 pt) In the table below, indicate the general function of the proteins encoded by each
gene class (e.g. transcription factor, receptor, etc.)
3c) ( 5pt)In the table below, indicate one phenotypic feature that is commonly observed in
fly larvae or adults that are homozygous for a loss-of-function mutation within each gene
class
Order of
Gene class
General function of Loss-of-function mutant phenotype in
action in fly
protein products
fly larva or adult
development
1
maternal effect
genes
transcription factor
lethal
larvae lacking polarity
larvae with mirror imaged anteriorposterior features
2
gap genes
transcription factor
lethal
larvae missing central features
mirror imaged remaining features
3
pair rule genes
transcription factor
lethal
failure to establish larval segments
4
segment polarity
genes
secreted peptides
receptors
transcription factor
(engrailed)
in whole embryo:
lethal
loss of larval segment polarity
loss of parasegment boundary
in mutant sectors:
missing adult structures
mirror-imaged adult structures
5
segment identity
genes
transcription factor
altered identity of adult segments
homeotic transformation of adult organs
3d) (2 pt)Which of the gene classes listed above are highly conserved in gene sequence
and gene product function, between insects and mammals?
Segment polarity and segment identity genes
3e) (3 pt) Which of the gene classes listed above have functions unique to drosophila and
related insects, and what developmental feature or advantage is conferred by those gene
functions? Maternal effect, gap and pair-rule genes. These allow for rapid establishment
of embryo patterning and therefore a rapid developmental program
PCB5065 Fall 2010
Exam 4
page 5 of 5
Name ____________________________________________
5. (20 pt.) The diagram below shows the ABC model describing the specification of different
floral organs in the four whorls of the arabidopsis flower. The spatial positioning of floral
organ identity gene (A, B and C) expression is shown, along with the resulting pattern
of floral organs.
A
B
B
A
C
C
sepal petal stamen carpel
A
A
C
C
sepal sepal carpel carpel
x (4 pt)
A
B
B
A
A
A
sepal petal petal sepal
y (5 pt)
C
B
B
C
C
carpel stamen stamen carpel
z (5 pt)
5a) (2 pt)Based upon the ABC model diagrammed above, indicate which gene or combination
of genes that acts to specify:
Sepals A alone
Petals A+B
Stamens B+C
Carpels C alone
5b) (1 pt) Which two gene classes in the diagram above mutually repress each other’s
expression ?
A and C
5c) Over x (above), use boxes to diagram the spatial distribution of remaining gene products if
B function is entirely lost by mutation; underneath your boxes, indicate the flower
structures that are predicted to develop. Just label them (e.g. sepal, petal, etc.); no
need to draw them.
5d) Over y (above) use boxes to diagram the spatial distribution of gene products and then
indicate the flower structures predicted to occur if C function is entirely lost by mutation.
5e) Over z (above) use boxes to diagram the spatial distribution of gene products and then
indicate the flower structure predicted to occur if A function is entirely lost by mutation.
5f) (3 pt) In the space below, briefly compare and contrast the nature of the plant floral organ
identity genes with the drosophila segment identity genes respect to their
evolutionary origins and functions. 1 pt each, any three; note the question specifically asked
about organ or segment identity, not about development in general
•
•
•
•
•
•
•
C
Both encode transcription factors
Drosophila, indeed all animals, use the homeobox transcription factors to specify segment or
organ identity.
Although plant genomes contain homeobox transcription factor genes, plants recruited a
different class of transcription factor genes, the MADS box genes, to specify organ identity
Evolved independently
Mutual repression observed in both systems
Mutants condition homeotic phenotypes in both systems
In animals the homeobox genes are clustered, whereas the plant MADS box genes are not