Download Name______KEY Genetics C3032 - Examination #2

Name__________KEY_________________
Genetics C3032 - Examination #2 - October 31, 2006
General instructions: Don't Panic. Be sure your name is on every page and that you write your exam in ink. Answer
the questions in the space provided. Clearly state your reasoning; if I can understand what you are saying during the
grading, there is a greater chance that you will get at least partial credit. The value of each question is indicated.
1. (24 points) A researcher working on Caenorhabditis elegans wants to map a dominant mutation d relative to two
known recessive mutations a and b that are 2 map units apart. Because the d mutation is dominant, he decides to
save some time and simply look to see whether the A and B animals among the progeny of ab/d heterozygotes also
have the D phenotype (he ignores the other animals). The results are:
19 A progeny
3 AD progeny
2 B
progeny
16 BD progeny
a. (8) Assuming that this researcher knew what he was doing, draw a genetic map showing the positions of all three
genes. Give a numerical map position, not just an indication of approximately where d maps relative to a and b.. Show
your calculations.
a 0.25 d 1.75 b
There are 5/40 recombinants in the ad interval (AD and B); 5/40 X 2 = 0.25
b. A coworker suggests that the research reconsider his results in light of the possibility that d may be far away from a
and b.
i. (3) What single-recombinant chromosomes would result if d is far to the left of a?
dab, +++, +a+, d+b
ii. (3) What single-recombinant chromosomes would result if d is far to the right of b?
abd, +++, a+d, +b+
iii. (4) Using the information from parts i and ii, indicate what recombinant phenotypes (there are two) could
the researcher have looked for in his original experiment that would have told him that d actually mapped far outside
the ab region? Explain your answer. Hint: these phenotypes would not have arisen easily if d was between a and b.
Wild type (+++/ab) and DAB (dab/ab). These only arise from double crossover events if d is between a and b.
iv. (6) If one assumes that d is not between a and b, is d to the right of b or to the left of a? Explain your
answer. (You should use the data given above in your answer, but you should not try to give a numerical value for the
position.)
Compare AD recombinant animals with A recombinant animals. When d is to the left, A animals arise from one
recombination event (+a+/+ab) and AD animals arise from two events (+a+/dab). In other words AD animals are
rarer than A animals. When d is to the right, A animals arise from two events (on the same chromosome:
+a+/+ab) and AD animals arise from one event (a+d/ab). A animals are rarer than AD animals. Since there are
more A than AD animals, d is to the left.
Name__________KEY_________________
2. (15 points – 5 each) A series of mutant lines in C. elegans were mated with several tester strains with the following
results. Provide an explanation for each of the possible abnormalities in the mutant lines.
a. Strain A (which was heterozygous for defect a) was mated to strain that was homozygous for recessive mutations in
seven closely linked genes (jklmnop). Half of the resulting cross progeny showed the L, M, N, and O phenotypes.
Strain A has a deletion that uncovers the l, m, n, and o genes. It fails to complement the recessive mutations
in these genes.
b. Strain B was mated to a strain homozygous for recessive mutations in two genes (q and r) that map to the second
and third chromosomes, respectively. The resulting heterozygotes were allowed to self, and or wild-type or QR animals
were seen.
Strain B contained a translocation causing pseudolinkage of the two unlinked mutations.
c. Strain C was mated to a strain homozygous for recessive mutations in two genes (s and t) that normally map five
map units apart and their progeny were allowed to self. None of the resulting F2 animals had only the S or the T
phenotype. When Strain C was mated to a strain homozygous for recessive mutations in two genes (q and r) that map
to the second and third chromosomes, respectively, and the resulting homozygous animals were allowed to self, Q, R,
wild-type, and QR animals were seen. (Strain C on its own looks wild type.)
Strain C probably has an inversion; this prevents recombination.
3. (20- 5 points each) Define and distinguish briefly between each set of the following pairs
a. PD/NPD ratio of linked vs. unlinked genes in yeast
If the genes are unlinked, then PD=NPD because of independent assortment; if the genes are linked, then
PD>>NPD, since the former results from segregation of nonrecombinant chromosomes and the latter arises
from a double crossover.
b. Profiling gene expression by mosaic analysis vs. profiling gene expression using lacZ or gfp promoter strains.
Mosaic analysis tells where a gene’s activity is needed; lacZ or gfp promoter strains indicate where a mRNA is
made.
c. DNA transposons like P factors vs. retroposons like Ty1.
DNA transposons excise and then insert into other regions of the DNA; retroposons make an RNA
intermediate.
d. Conversion vs. recombination using Hfr and F- strains.
Conversion from an F- to F+ is rare because the F factor enters the F- cell late; recombination occurs much
more frequently because the host chromosome is transferred first (hence the name Hfr).
Name__________KEY_________________
4. (25 points - 5 points each) Explain the following:
a. How duplication and deletion strains can be generated from translocation strains in Drosophila.
Translocations have parts of chromosomes. By choosing different translocations so that they all but a small
part of a chromosome, the resulting organism will be deleted for that region. By choosing translocations with
small overlaps, one can produce duplications.
b. How replica plating can be used to isolate new Hfr strains.
F+ E. coli defective in some genes and wild type for others are plated on enriched medium. They are replicated
plated onto a plate with minimal medium that is uniform with F - coli that are wild type for the missing genes
but lacking others found in the F+ bacteria. Cells will grow where an Hfr allowed mating. Go back to the
original plate and select bacteria from the equivalent area.
c. How Hfr strains can be used to determine map order E. coli.
Measure positions by the timing of entry of the Hfr chromosome
d. How Hfr strains can be used to determine map order E. coli in a different way.
Hfr strains allow entry of DNA for recombination map order can be determined by doing reciprocal crosses.
e. How transposons can be used to identify genes encoding eye components in Drosophila.
Enhancer trap: insertion allows expression of -galactosidase or GFP, then insertion can be cloned.
Name__________KEY_________________
5. (18 points) Prior to meiosis a Neurospora diploid is heterozygous for five genes on one of its chromosomes;
specifically it has genotype ABCDE/abcde. During meiosis the following four-strand intermediate resulted from DoubleStrand Break Repair (note that other strands involved in meiosis are not drawn).
5’
’
5’
A
B
c
d
E
A
b
c
d
E
a
B
c
D
e
a
b
c
d
e
5’
5’
a. (4) On which chromosome did the double-strand break occur (ABCDE or abcde); where on the chromosome did the
break occur; and what was the consequence of the break?
ABCDE chromosome at the C gene. As a result of the break and loss of the C gene, it has been replaced by
the c gene.
b. (4) Assume that the genes are equally spaced along the chromosome. What evidence suggests that unequal
resection (loss of DNA) occurs?
The D gene has been lost and replaced by d. The same has not happened with B.
c. (4) Diagram the octads that would result from resolution (break and reformation) of the left Holliday Junction as
shown (i.e. horizontal resolution), and of the right Holliday Junction by either horizontal or vertical resolution. Assume
that mismatch repair does not occur. Which, if either, octad shows evidence of recombination?
Horizontal-Horizontal
ABCDE
ABCDE
ABcdE
AbcdE
aBcDe
abcde
abcde
abcde
Horizontal-Vertical
ABCDE
ABCDE
ABcde
Abcde
aBcDE
abcdE
abcde
abcde
Recombination is seen with H-V
d. (4) Draw the same octads assuming that mismatch repair is 100% efficient and changes any mismatches so that
complete gene conversion (i.e., to unexpected allele) results.
Horizontal-Horizontal
ABCDE
ABCDE
AbcdE
AbcdE
aBcDe
aBcDe
abcde
abcde
Horizontal-Vertical
ABCDE
ABCDE
Abcde
Abcde
aBcDE
aBcDE
abcde
abcde