Advanced Genetics Key 2011

advertisement
Gloria A. Moore
Page1
2/6/2016
Name_______________________________ (Initial each page)
PBC5065 Advanced Genetics – Moore Section 2011 Test – 10/24/11
Show your work, or write me notes! Partial credit may be given if your answer is incorrect
but I can follow your logic.
Hirsuta Much and her eager assistant Igor do slightly mad but sound science in Hirsuta’s wellequipped lab in Transylvania. Igor, in one of his late-night ramblings, has captured an extremely
hairy male mouse and brought it back to the lab. He thinks it would make a nice pet or a tasty
snack, but Hirsuta can see that the mouse has scientific potential if its phenotype is due to a
simple genetic alteration. Hirsuta has obtained grant funding from the International Society of
Hair Restoration Surgery, The Trichological Society, and the Luxuriant, Flowing Hair Club for
Scientists to study a rare condition in humans that causes long, thick hair to grow all over their
face and bodies. Although the condition is rare, Hirsuta has identified several families in which it
occurs. The pedigrees of the families point to a single gene as the cause of the phenotype.
Hirsuta is calling the putative gene Werewolf. Could the phenotype of the hairy mouse be due to
a similar gene?
Gloria A. Moore
Page2
2/6/2016
Name_______________________________ (Initial each page)
To determine whether this is the case, Hirsuta mates the hairy male mouse (W) to two normal
female mice (N), each of whom produces 10 offspring. The results are shown in the figure
below.
Phenotypes
N W N
W N N N W W W W N W
1 2
3
4
5
W N N W N W N N
N W
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
2
0
21
22
23
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
1) After seeing the results of the crosses, Hirsuta hypothesizes that the phenotype of the hairy
dad mouse (number 12 in the figure) does appear to be due to a simple genetic alteration (let
us call it WEREWOLF). Assign allele designations for WEREWOLF.
What is the genotype of hairy dad at the WEREWOLF locus? (2 pts)__Ww, or Wn, or
something designating heterozygosity_________
What is the genotype of both mom mice (numbers 11 and 13)? (2 pts) ___ww, or nn_or
something designating homozygosity ________
I decided you could have these genotypes reversed and still get a correct answer. If you
designated one homozygote and one heterozygote, I gave full credit.
Hirsuta uses 200 markers to genotype all of the mice. The results for 3 markers are shown in the
figure above..
Gloria A. Moore
Page3
2/6/2016
Name_______________________________ (Initial each page)
2) (2 pts) Which marker is dominant in nature? C is the correct answer; A and B are codominant
3) (4 pts) Which marker or markers appear to be linked to WEREWOLF? _A and B_ .
4) Which allele for Marker A is associated with the WEREWOLF phenotype? (2 pts) __
WEREWOLF phenotype and upper band allele of Marker A
Which allele for Marker B is associated with the WEREWOLF phenotype (2pts)?
WEREWOLF phenotype and lowest band of Marker B (Tell me what allele designations
you are using) I gave partial credit if you seemed to have some knowledge of the concept
even if you were not correct.
5) (4 pts) Circle the progeny mice that appear to be recombinant between WEREWOLF and any
of the markers (indicate which marker)
Progeny 10 & 20 are recombinant for Marker A
Progeny 20 is also recombinant for Marker B.
With this kind of segregating progeny, it is possible to test for linkage using either Chi Square
analysis or LOD score analysis.
6) Set up a Chi square analysis to test for linkage between Marker A and WEREWOLF (i.e.
show the observed and expected categories and numbers for each class and the degrees of
freedom for the test). (4 pts) (Marker A alleles lettered from top A and B.)
Genotype
WwAB
WwBB
wwAB
wwBB
Totals
Df = 3
Observed
9
1
1
9
20
Chi Square Table
df 0.995
0.99
----1
2 0.010 0.020
3 0.072 0.115
4 0.207 0.297
5 0.412 0.554
0.975
0.001
0.051
0.216
0.484
0.831
(O – E)2/E
3.2
3.2
3.2
3.2
12.6 = X2
Expected
5
5
5
5
20
0.95
0.004
0.103
0.352
0.711
1.145
0.90
0.016
0.211
0.584
1.064
1.610
0.10
0.05 0.025
0.01
2.706 3.841 5.024 6.635
4.605 5.991 7.378 9.210
6.251 7.815 9.348 11.345
7.779 9.488 11.143 13.277
9.236 11.070 12.833 15.086
0.005
7.879
10.597
12.838
14.860
16.750
Gloria A. Moore
Page4
2/6/2016
Name_______________________________ (Initial each page)
Does the Chi square value appear to be significant or nonsignificant? What does this indicate?
(4 pts) Significant. Alleles at genes not assorting independently (I counted off one point if
you said genes were linked and didn’t mention independent assortment.)
7) (2 pts) Which marker appears to be most closely linked to WEREWOLF? Why?
_____B________
Would this be a good marker from which to initiate a chromosome walk to WEREWOLF? Why
or why not? (2 pts) I gave one point of credit if you said yes, but justified why. I gave 2 pts
for realizing that, based on the recombination frequency in this one small progeny, this
would be a very long way to walk, and that closer markers are needed.
Hirsuta next searches for more markers that are linked to the marker you selected in the last
question, reasoning that she does not need mice segregating for WEREWOLF to locate these
markers, but if new markers that are linked to the already identified marker are found, at least
some of them should also be linked to WEREWOLF. Therefore, she analyzes a number of
mouse families segregating for alleles at the selected marker for linkage to other markers. A table
of the data is shown below.
Theta value
Number of 0.05
0.1
0.2
0.3
0.4
informative
families
Tgf6
4
1.20 2.30 3.40 3.10 2.70
B1123
6
4.71 10.55 5.16 2.23 0.20
R99
4
0.75 1.82 3.85 4.10 2.42
MouseAB13 5
6.41 4.28 3.05 1.20 0.05
M13-687
2
1.22 2.03 1.33 0.71 0.35
M13-233
4
1.21 1.76 3.04 1.65 0.45
M14-112
2
0.91 0.98 1.35 1.45 1.10
Sa11
3
-3.21 -1.11 -1.02 -0.45 -0.01
Lx55
3
1.90 1.20 0.80 0.20 0.05
H12-9863
5
-1.45 0.01 0.12 2.44 2.20
Marker
Locus
8) (2 pts each)
Which of the new markers appears to be most closely linked to your selected marker? Why?
MouseAB13, it has the highest LOD at the lowest Theta value shown.
Does this mean that it is more closely linked to WEREWOLF?
No, this marker could be on the other side of your marker from WEREWOLF.
Which new marker or markers clearly do not show linkage with your selected marker?
Gloria A. Moore
Page5
2/6/2016
Name_______________________________ (Initial each page)
After grading a number of papers, I gave full credit for Sal1 alone and if you listed
other markers with highest LOD scores at high thetes, because I think Jim and I have
different views on this.
After a lot of linkage analysis, Hirsuta identified a marker, Marker X, that appeared to be very
tightly linked to WEREWOLF. Using a rat/ mouse somatic hybrid chromosome panel (where
mouse chromosomes are eliminated), Hirsuta was able to map Marker X to mouse chromosome
6. Hirsuta also had access to a set of radiation hybrid cell lines for mouse (made by irradiating
mouse chromosomes and then fusing them with hamster chromosomes). The part of the panel
that includes mouse chromosome 6 is shown on the next page. The individual hybrid lines are
indicated by their names across the top of the figure and numbers 1 to 30 across the bottom of
the figure. The portion of mouse chromosome 6 that each hybrid cell line harbors is shown by
the line running vertically between the name and number. Individual regions of chromosome 6
based on Giemsa banding are designated at the far left of the figure by numbers.
Hirsuta extracts DNA from each of the hybrid cell lines and probes it with a probe made from
Marker X. Individual hybrid lines 1, 2, 4, 6, 7, 8, 10 13, 15, 17, 19 21, 23, 25, 26, 27, 28, 29, and
30 contain mouse DNA that hybridizes with Marker X.
Gloria A. Moore
Page6
2/6/2016
Name_______________________________ (Initial each page)
9) In which region is Marker X and so presumably WEREWOLF, located? (4 pts) Band 36,
hybrid 30 is diagnostic for this.
Would the gene be on the same chromosome in the human genome? (4 pts) No,
rearrangement, speciation, translocations.
Thanks to the genetic resources available for mouse, Hirsuta is able to order large insert clones
that cover the portion of chromosome 6 that appears to contain WEREWOLF.
10) (2 pts) Name two types of large insert clone that she could potentially use. YACs, PACs,
BACs
The inserts in the clones have already been put in order in the mouse genome, and some markers
have been identified on each clone, so Hirsuta could do more linkage mapping to move still
closer to WEREWOLF. However, since she now has a relatively small part of the genome to
search, she decides to search for candidate genes in the region. After still more experiments and
examination of the available databases, she identifies three transcripts in the region that she
believes might have potential to be WEREWOLF.
One of the things Hirsuta does next is prepare RNA from various mouse tissues and assemble
them on northern blots, that she then hybridizes with probes from each individual interesting
transcript (i.e. potential gene of interest, WEREWOLF). The blots are shown below.
Candidate gene 1
→
Candidate gene 2
→
Candidate gene 3
→
Which candidate gene is most likely to be WEREWOLF? Why? (2 pts)
Gloria A. Moore
Page7
2/6/2016
Name_______________________________ (Initial each page)
Candidate 1, because the gene is expressed only in skin, which makes sense for Werewolf._
What would you expect to see if you probed a zoo blot with WEREWOLF (or did a homology
search at NCBI)? (2 pts) Full credit given if you said something indicating that a
homologous gene might be present in mammals but not other species.
Igor has found a second mutant mouse
that has long hair (top mouse in figure)
compared to his littermate (bottom
mouse). Using a Werewolf probe and
FISH technology, Igor examined the
configuration of Werewolf in the mutant
mouse and its parents. His results are
shown below. The DNA represents
mouse chromosome 6 and the arrows
represent copies of Werewolf.
The differences in gene number in the three mice are an example of what phenomenon we
discussed in class? (2 pts) I accepted aneuploidy, copy number variation, duplication. The
point of the example is that it is copy number, not copy arrangement, that leads to the
mutant phenotype.
The mutant mouse displays what cytological aberration if all of the rest of his chromosomes are
normal? (2 pts) Trisomy for chromosome 6.
Gloria A. Moore
Page8
2/6/2016
Name_______________________________ (Initial each page)
In addition, Igor has located a mouse genetic
stock which is segregating for mice with no
hair. In the diagram below, a litter of 4 mice and
their parents are shown, along with Igor’s
cytological analysis. The naked mice are
represented by dark symbols, chromosome 6 is
represented by the DNA, Werewolf copies are
represented by arrows, and the X indicates that
the gene region is missing, not just defective.
Gloria A. Moore
Page9
2/6/2016
Name_______________________________ (Initial each page)
The naked mother and daughter display what cytological aberration? (2 pts) _Deletion____
What has to happen, compared to the other mice in the pedigree, for the affected mice to be
naked? (2 pts)
They have to have one (or fewer?) copies of the gene, no matter the configuration of the
genes if there is more than one copy (again copy number variation).
Igor loves to do cytogenetics, especially with chromosome paints (FISH) (he thinks the resulting
preparations are pretty). He prepares a number of mouse cell lines for Hirsuta to use in her
studies of WEREWOLF expression, and several different kinds of FISH paint probes. Match the
type of paint probe Igor prepared with the experiment Hirsuta did with it (some experiments can
be done using more than one kind of probe and in some cases it would be useful to use more than
one kind of probe) (see next page):
Gloria A. Moore
Page10
2/6/2016
Name_______________________________ (Initial each page)
Probes:
A. A probe composed of repetitive DNA (SINEs and LINEs)
B. FISH with a whole chromosome library, where each chromosome is differently colored
C. A probe specific for all of chromosome 6
D. A probe for a small region around WEREWOLF
E. A probe for telomeres
F. A probe of the alphoid satellite sequence specific to chromosome 6
Experiments: (1 pt each)
__B___ Determining the chromosomes with which chromosome 6 most often interacts in an
interphase nucleus
C, D, F Studying WEREWOLF expression in a cell line trisomic for chromosome 6
_C, D_ Determining the location of WEREWOLF in the chromosome 6 interphase territory,
using 2 probes
__C__ Studying cell lines with duplications or deletions of chromosome 6 (assuming they are
large enough), or translocations involving chromosome 6, but not inversions in chromosome 6,
using mitotic metaphase preparations
__D__ Confirming where on chromosome 6 WEREWOLF is located, using mitotic metaphase
preparations
__A__ An experiment where all of the chromosomes are colored the same color
__E___ An experiment where the ends of all of the mitotic chromosomes are studied
__E, F_ Study of a cell line where there is an inversion on one homologous chromosome 6,
using 2 probes
There are other possible answers here. I gave credit for anything feasible.
So, after much effort, Hirsuta believes that she has identified the gene for WEREWOLF. She
has begun to characterize the gene in hairy and normal mice. Will alterations in the gene affect
baldness? Will this lead to a “cure” for baldness and make Hirsuta and Igor hugely rich?
Unfortunately, it doesn’t look likely. As Hirsuta looks at the effect of altering the sequence
and/or expression of WEREWOLF in normal mice, she observes deleterious alterations in the
phenotypes of characteristics other than hair production.
Gloria A. Moore
Page11
2/6/2016
Name_______________________________ (Initial each page)
The possible reason for this is (4 pts):
A. WEREWOLF is alternately transcribed in different tissues or at different
developmentally stages, and serves different purposes in those situations.
B. Exons of WEREWOLF are shuffled to produce different proteins.
C. Different regulatory agents influence WEREWOLF under different circumstances.
D. WEREWOLF may moonlight.
E. Any of the above may be true.
Most people got this correct. I also gave full credit for A,B, and C, since many
people didn’t know what moonlighting is. I gave partial credit for other answers for
the same reason.
Strangely enough, Igor is working with a similar gene in the model plant Arabidopsis. Of course,
plants don’t have hair, but some of them do have trichomes.
Igor has identified a mutant where the trichomes have more lobes than usual: The mutation
appears to be due to a single gene. To tease Hirsuta, Igor has also called this gene Werewolf.
Further, Igor used a similar strategy to positionally clone his gene, by mapping the Werewolf
gene location compared to those of molecular markers scattered throughout the Arabidopsis
genome. Like Hirsuta, Igor has localized his Werewolf gene to a particular part of the
Arabidopsis genome using physical mapping methods. There are four candidate genes in the
region. Igor clones the promoter region of each candidate gene and inserts each one individually
into a genetic transformation vector plasmid in front of the GUS gene. GUS is a scorable marker.
When it is being expressed, the tissue where it is being expressed stains blue. Igor then makes
transgenic plants using each of his constructs and stains them for GUS. The phenotypes of the
four transgenic types are shown below.
Circle the
plant transformed with the gene promoter that is most likely to be that of Werewolf. (4 pts)
Gloria A. Moore
Page12
2/6/2016
Name_______________________________ (Initial each page)
Although Igor is proficient at many techniques in the lab, he has not had any formal training in
genetics. Hirsuta is trying to teach him basic concepts. She asks him questions and rewards him
with a tasty treat when he answers correctly, usually an insect (she draws the line at mice).
11) (2 pts) All of the following proteins are present in the nucleosome core except (circle letter of
correct answer):
A. H1.
B. H2A.
C. H2B.
D. H3.
E. H4.
12) (2 pts) Which one of the following types or classes of DNA sequences comprises the largest
portion of the human genome?
A.
B.
C.
D.
E.
Protein-encoding sequences (exons)
Single-copy sequences that do not encode proteins
Intron sequences
Dispersed repetitive DNA (SINEs and LINEs)
Sequences encoding ribosomal RNA
13) Name one type of marker you would use for mapping in mouse if there were no sequence
available from the organism.
RAPD, AFLP, RFLP
14) True or false: (2 pts each)
__F___The histone proteins in nucleosomes have only a structural function.
__F___Highly repetitive DNA elements such as Alu sequences in humans are never
found within any part of the DNA sequences of genes that code for proteins.
__T___Some species tolerate polyploidy well, but aneuploidy is detrimental to almost all
species.
__T__Expanding repeats of short DNA repetitive sequences may be a causal factor for
some diseases or syndromes.
__T__Telomerase function is implicated in aging, cancer, and serves other important
cellular activities.
__F__ There are 80,000 to 100,000 protein-coding genes in the human genome.
Gloria A. Moore
Page13
Name_______________________________ (Initial each page)
2/6/2016
Download