Advanced Genetics – Moore/ Olmsted Exam – Monday, October 22, 2012 - KEY
Name_______________________
Initial all pages. Show your work! We give partial credit, but only when we can tell what
your intentions are.
Scary Larry, an evil plant breeding genius wanted to develop Venus flytrap plants that were large
enough to eat humans. On a collecting trip, he found a plant that looked to have a large enough
mouth to meet his needs.
Dr. Larry wanted to map the location of the “Big Mouth” trait. The genomic resources available
for venus flytrap are an expressed sequence tag (EST) library and some Next Generation
sequencing reads from various potential parents that he might use in crossing.
What would be the best marker type for Dr. Larry to use and why? (4pts)
Full credit for picking PCR-based markers that require some prior sequence information
(because of the EST library and NGS) like SSR or SNP. Partial credit for other answers as long
as you gave a marker type and a rationale for why you would use it
What are three characteristics of “good” DNA markers? (3pts)
Pick from: High polymorphism, codominant, locus specific, frequent distribution across the
genome, easily detected, high repeatability, scalable, low development/operational costs
Why would he not want to use RAPD markers? (2pts)
I was looking for lack of repeatability, but also accepted not necessary because the sequence
information, and I gave 1 point if it sounded like you knew what a RAPD marker was.
If he found a marker in the actual “Big Mouth” gene, what would we refer to that marker as?
(1pt)
Functional marker or perfect marker
Scary Larry made a cross between the homozygous plant he collected and a homozygous
accession that he had already been working with. He then self-pollinated an F1 individual to
create an F2 population of 10 individuals. From these 10 individuals, he collected the following
data for the “Big Mouth” trait (we’ll use the Bm designation for the gene) and three molecular
markers:
Parents
F2 Progeny
Wild
"Big Mouth"
phenotype
Mutant
type
F1
1
2
3
4
5
6
7
8
9
10
Yes
No
Yes
Yes
Yes
No
No
Yes
Yes
Yes
No
Yes
Yes
Marker 1
Marker 2
Marker 3
Which marker is not useful to you for linkage mapping and why? (2pts)
Marker 1 because it does not segregate
Which marker is co-dominant and why? (2pts)
Marker 3 because you can see both parental bands in the F1 heterozygote
For the Bm and marker 2, how many recombinant individuals are there? (1pt)
1 – Individual 5 because it has the Big Mouth phenotype but lacks the fragment present in the
mutant
For the Bm trait and marker 3, how many recombinant individuals are there? (1pt)
3 – Individuals 1, 4, and 9 because recombination between the top fragment and the Big Mouth
phenotype
.
What is the recombination frequency between Bm and marker 3? (2pts)
3 recombinants / 10 total = 0.3. I gave full credit for 3/10, .3, or 30cM (even though I didn’t ask
for linkage distance. I also gave full credit if you got the number of recombinants wrong from
the previous question but had the right idea for how to determine recombination frequency.
What is the relationship between recombination frequency and linkage distance and how do we
report linkage distance? (4pts)
The relationship is that linkage distance is calculated based on recombination frequency. Fewer
recombinants (lower recombination frequency) results in smaller linkage distance. I tried to give
credit if you explained this in words, diagrams, or numbers, but the idea that linkage distance is
based on recombination frequency had to be there. Distances are reported in centimorgans (cM).
How is physical distance related to linkage distance? (2pts)
Something about loosely related, not linearly related, or an explanation of how double
crossovers, recombination hotspots, centromeres, etc. can increase or decrease recombination.
Would you consider marker 3 and Bm to be closely linked (1pt)
They are linked, but not closely at 30 cM
How would a double recombinant impact his estimation of linkage distance (1pt)
Double recombinant would lead to underestimation of linkage distance. I gave credit if you
explained or drew the process.
Do you think venus flytrap is a diploid or polyploid plant and why? (2pts)
The fragment patterns on the gel image on the second page indicate these plants are diploid.
Some of you also used the karyotype below to deduce it is diploid. I also gave credit for the
conclusion that the plant is polyploid based on the large organ size (big mouth).
Scary Larry decided to improve his developing linkage map by sequencing his two parents to
identify single nucleotide polymorphism (SNP) markers. He now has a lot of SNP markers but
is finding that he has not been able to reduce the linkage distance between markers substantially.
What might be an explanation for his inability to find closely linked markers? (4pts)
For full credit, what I was looking for was recognition that with a population of n=10, there are
not enough potential recombinants to be able to find very closely linked markers. I tried to be
generous with points for many potential reasons why closely linked markers might not be found,
like recombination suppression, centromere region, etc.
In addition to the Bm mutant that he found, Scary Larry had a greenhouse full of venus flytrap
mutants. To better characterize these mutants Dr. Larry did some cytogenetics and flow
Cytometry to determine chromosome number. The karyotype for his control venus flytrap
looked like this:
Is the control a diploid or polyploid? (1pt)
Diploid
There was one very strange looking plant in the greenhouse. It appeared to be sterile and did not
produce progeny when crossed with other venus flytrap parents.
Would we call this individual an aneuploid or euploid? (1pt)
Aneuploid
How are polyploids different from aneuploids? (2pts)
Polyploids have increases in complete sets of chromosomes, aneuploids have variation in
chromosome numbers (plus or minus) that is not the exact multiple of the haploid number
What might be an advantage of polyploidy? (2pts)
I talked about heterosis, gene redundancy, and loss of self-compatibility in class, but there are
many potential answers here.
Harry Butz is a human geneticist studying colon cancer in humans. Harry believes that there is a
gene responsible for the late on-set formation of cancerous colon tumors in some individuals that
segregates in a dominant autosomal fashion.
Harry decides to try to clone the gene responsible for the colon tumors and study it at a
molecular level. He has investigated a number of markers for linkage to the colon cancer trait by
looking at many affected human families. Some of the information he has obtained is
summarized in the table below.
Marker
Locus
Rect 1123
Rect 1314
Chyme 66
Pol 457
Pol 772
Fec 4537
Chyme 12
Pol 507
Fec 2201
Pol 333
Number of 0.01
informative
families
3
-3.21
4
-2.00
4
0.09
4
-0.08
6
5.90
2
1.38
5
9.71
2
-0.04
3
2.40
5
-1.90
0.05
-2.21
-1.55
1.21
0.85
10.20
1.45
7.21
0.91
1.90
-1.45
Theta Values
0.1
0.2
-1.11
0.05
3.04
1.62
4.80
1.33
5.28
0.98
1.20
0.01
-1.02
0.17
1.76
3.85
2.80
1.21
3.16
1.35
0.80
0.12
0.3
-0.45
1.44
1.65
1.80
1.20
0.56
2.23
1.45
0.20
2.44
Which marker do you think Harry should use to initiate a physical walk (i.e. for positional
cloning) toward the colon cancer gene? Why? (4 pts)
Chyme 12. Few people missed this; I harped on it a lot in class. I gave one point if you were
totally wrong, but had a rationale.
Which markers do not appear to be linked to the colon cancer gene? Why? (2pts)
To get both points, you had to realize that the markers had negative LOD values or that they had
values less than 3 (it was OK to say the latter were inconclusive).
Harry is also studying this disease in a mouse model. Here he has obtained an inbred mouse line
that is segregating for what he thinks may be the same tumor-causing gene and a marker, with
alleles AB. In one backcross mouse population, Harry obtains the following results (C =
dominant colon cancer allele, c = wild type allele; AB = marker with alleles A and B):
Genotype
CcAB
CcBB
ccAB
ccBB
Totals
Observed
42
12
5
35
94
Expected
23.5
23.5
23.5
23.5
94
Chi square
14.56
5.63
5.63
14.46
40.38
P value (df)
---------------------
a. Is this a significant Chi square value? At what value? (2 pts) Yes, at 0.005.
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
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
b. What would Harry’s conclusion be from this result? (4 pts)
You did not get full credit if you only said the loci were linked. There proper thing the test tells
you is that alleles at the loci are not assorting independently, implying that the loci may linked.
That you can differentiate alleles from loci is really important to me.
c. What is your estimate of recombination distance from this data? (You can just put the
fraction if you want.) (2 pts)
17/94
Before Harry begins his molecular cloning studies, he wants to assemble the materials he will
need. He also will use a mouse model system to help isolate and characterize the gene. Which of
the following materials are likely to be helpful to him? (Circle helpful ones.) (6pts)
a. A cDNA library made from the liver of a human fetus.
b. An arrayed human genomic library in a BAC vector made from flow sorted
chromosomes.
c. A mouse/ hamster somatic hybrid mapping panel.
d. A preparation of rat brain RNA.
e. A cDNA library made from adult human colon cells.
f. A cDNA library made from colon tumor cells.
B, C, E, F. I deducted 2 points if you didn’t identify all of these and you didn’t have a rationale.
Harry narrowed down the human genomic region where he thought his gene was located (he
could still be far away). When he compared the locations of known genes on the linkage map of
the region with the same genes on a physical map of the same region, he observed the following:
The maps are not identical because (3 pts):
a. There is no relationship between the position of the genes in a genetic map and
their positions on the DNA.
b. Recombination frequencies per kb of DNA are not uniform throughout a
chromosome.
c. The further apart two genes are, the more likely they are to recombine.
d. The closer two genes are, the more likely they are to recombine.
e. Some genes contain introns.
B. C and E are true, but don’t answer the question. However, I decided it was
reasonable to look at recombination as described in C, so I didn’t count off if you
identified only B and C.
By chromosome walking, Harry has reached the area of the human genome wherein he thinks his
gene of interest lies. The region is shown in the figure below. Unfortunately, there are at least
three apparent genes in the region. Label the following in the figure: a BAC clone (or partial
BAC clone); a gene; an intron in a gene; an exon in a gene. (4 pts)
Most everybody got this correct. I allowed some latitude on BAC clones.
Harry then tests the human genes in the region he has identified to determine which one might be
the best candidate for the adult onset colon cancer gene. One thing he does is northern analysis.
From his analysis, which gene do you think is the best candidate? Why? (6 pts)
HTRE3; in adult colon and intestine. If you said
HTR3D, you only got partial credit. I told you the disease was late onset so you wouldn’t expect
the gene to be expressed in fetal tissue (or in liver or spleen).
Harry used his newly identified human gene (let’s call it ColC, which mapped to human chromosome 3 –
This was a mistake on my part – see below) as a probe on his mouse/hamster somatic cell line hybrid
mapping panel. Shown below are part of his results using four DNA clones to probe mouse/hamster hybrid
cell tissue culture lines a-f, where + means the probe hybridized to a Southern blot of the DNA from that
cell line. The mouse chromosomes present in each cell line are indicated by a + in the second part of the
table. (8 pts)
CELL LINES
a b c d e f
Probe for ColC
++- - - -
Probe for VB6
- ++- ++
Probe for RJ776
++- - - +
Probe for KLL89
++++++
Mouse chromosome 1 + + - - - +
Mouse chromosome 4 + + - - - Mouse chromosome 17 + + + + + +
Mouse chromosome 22 - + + - + +
What is the chromosomal location of gene: ColC?__4___ VB6?__22___ RJ776? __1___ KLL89? __17___
To Harry’s surprise, ColC did not map to mouse chromosome 1. Luckily, Harry is married to the brilliant
cytogeneticist Ona Thera. She produced a comparative map of human and mouse chromosomes as follows:
a. What techniques were used to produce this information? (4pts)
I accepted pretty much anything reasonable here, but if you just said chromosome painting or
FISH alone, you did not get full credit.
b. What is the primary cytological event that probably led to the differences between the genic
arrangements on human and mouse chromosomes? (4pts)
I was expecting you to say translocation but again I accepted anything reasonable. I did not
accept speciation, since I specified a cytological event.
c. On which mouse chromosome did the gene most likely map (3 pts)?
Everyone should have full credit for this, no matter what you said, because I made a mistake
above that said the human gene was on chromosome 3 when it should have said chromosome 1.
If it had said chromosome 1, I was expecting you to say mouse chromosome 4.
Ona offered to do some cytological investigations for Harry using his ColC probe, with other probes as
controls. Their idea was to look at normal cells and tumor cells to see whether there were cytogenetic
differences in normal and tumor cells and in the presence of ColC.
The following chromosome painting (FISH) probes were made: (1) one complementary to ColC; (2) one to
a highly repeated sequence such as Alu; (3) one to the entire mouse chromosome on which ColC mapped;
(4) one to DudD, a nontranscribed gene that is not located on the chromosome where ColC mapped; (5) one
to alphoid satellite sequences.
In an interphase chromosome preparation, which of the probes would hybridize to each of the following
regions? (write the probe name after each region) (5 pts)
a. a heterochromatic area within each territory, probably not on the periphery 5
b. to two places, possibly in heterochromatic regions away from the periphery, in territories far from
each other 4
c. all throughout all of the territories 2
d. to two entire territories, far from each other 3
e. to the periphery of two territories far from each other, or to loops extending from the territories 1
In a mitotic preparation, which of the probes would hybridize to each of the following regions? (write the
probe name after each region) (5 pts)
a. the centromeres of all of the chromosomes 5
b. all along all of the chromosomes 2
c. to a single spot on each of the homologous chromosomes where ColC mapped 1
d. to a single spot on two chromosomes somewhere else in the genome 4
e. to all of two chromosomes 3