Genetics Final Exam
100% Correct
Questions & Answers
, The cDNA prepared from cancer cells is labeled red and the cDNA prepared from normal
tissue is labeled green. A DNA microarray experiment is performed, and it indicates that
there are increased levels of expression of the gene dbl in the cancer cells compared to the
normal tissue. What result indicated this?
A) The spot corresponding to dbl is yellow.
B) The spot corresponding to dbl is red.
C) The spot corresponding to dbl is green.
D) The spot corresponding to dbl is black.
E) None of the above - ANSWER B) The spot corresponding to dbl is red.
2) Bicoid (B) is a maternal effect gene. Loss-of-function alleles (b) of bicoid are recessive. In a
cross between a heterozygous wildtype female (Bb) and a homozygous recessive male (bb),
which of the following is expected in the F1 progeny?
A) No progeny; the bb male will be sterile
B) Bb and bb adult flies
C) BB, Bb, and bb adult flies
D) Bb adult flies only as the bb embryos are defective
E) BB and Bb adult flies as the bb embryos are defective - ANSWER B) Bb and bb adult flies
TetR binds to DNA in the absence of aTc. LacI binds to DNA in the absence of lactose. In
this circuit (Circuit 2) all gene expression is constitutive (ON) unless repressed. In which
condition(s) will EYFP expression be OFF?
A) no aTc, no lactose
B) no aTc, added lactose
C) added aTc, added lactose
D) added aTc, no lactose
, E) A and D - ANSWER D) added aTc, no lactose
Which of the following would give you information about how total gene expression is
different in cells under heat shock conditions compared to normal conditions?
A) Compare cells grown in the two conditions using genome sequencing.
B) Compare cells grown in the two conditions by doing an immunoprecipitation
experiment with antibodies to a heat shock protein.
C) Compare cells grown in the two conditions using a DNA microarray
experiment.
D) B and C.
E) All of the above. - ANSWER C) Compare cells grown in the two conditions using a DNA microarray
experiment.
When cloning a DNA fragment that was cut with a restriction enzyme, the "sticky ends"
serve what function?
A) They base pair with each other and keep the two ends close so that DNA
ligase can join them
B) They keep the ends from being modified by DNA ligase
C) They ensure rapid degradation of the fragment by DNA ligase
D) They serve as the basis for recombination
E) They prevent ends generated by the same enzyme from being linked together - ANSWER A) They base
pair with each other and keep the two ends close so that DNA
ligase can join them
You have a particular Drosophila mutant with both homozygous males and females. When
you cross homozygous mutant males with wildtype females, all of the embryos develop
normally. Crossing the homozygous mutant females to wild-type males results in all of
the embryos dying, however. What is the best explanation for your observations?
A) The mutation affects a zygotic gene that's required during embryogenesis
, B) The mutation is on the Y chromosome
C) The mutation is a homeotic mutation
D) The mutation is on the X chromosome
E) The mutation affects the mother's ability to make the oocyte correctly - ANSWER E) The mutation
affects the mother's ability to make the oocyte correctly
You digest the plasmid pBBR322 (right) with EcoRI,
BamHI, and HindIII. What sizes are the resulting DNA
fragments?
A) 29, 375, and 4359 bp
B) 29, 346 and 3984 bp
C) 2, 29, 375 and 4359 bp
D) 159, 346 and 4174 bp
E) 31, 346, and 3984 bp - ANSWER E) 31, 346, and 3984 bp
In C. elegans, LAG-2 is a signal molecule that tells either Z1.ppp or Z4.aaa to not be an
anchor cell, and LIN-12 is the receptor that receives that signal on each cell. In two
different experiments, mutant worms were examined for the number of anchor cells. In
the first experiment, the worms were homozygous lag-2 (-/-). In the second experiment,
worms were heterozygous lin-12 (hyperactive/+). How do the expected phenotypes of these
experiments compare to each other?
A) The lin-12 (hyperactive/+) worms will have two anchor cells, while the lag-2 (-/-)
worms will have none.
B) The lin-12 (hyperactive/+) worms will have no anchor cells, while the lag-2 (-
/-) worms will have two anchor cells.
C) Both worms will have the same phenotype - neither Z1.ppp nor Z4.aaa will be
anchor cells.
D) Both worms will have the same phenotype - Z1.ppp and Z4.aaa will both be anchor
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