Presentation over the research I did at the University of Missouri of the summer. Sorry about the vagueness at parts, but I don't like to just read ppts slide for slide. I was speaking about the research with this in the background.

1. Blake Stephens

2. • Missouri Academy of
Science Mathematics
and Computing
• Northwest Missouri State
University
• Missouri Academy Graduate with:
Research Summer 1. High School Diploma from
Plattsburg High School
(MARS)
• Deanna Lankford 2. Associates Degree of
Science from Northwest
Missouri State University

3. • ATP plays an
important role as
an extracellular
signaling molecule.
• Purinergic
receptors are
responsible for
extracellular ATP
responses in
mammals, but
these same
receptors seem to
be lacking from
plant genome.

4. ATP stimulates closure of the Venus flytrap
It is apparent that
similar responses occur
in plants.
Jaffe (1973) Plant Physiol 51:17-18

5. To contribute to the discovery and categorization of
the plant ATP response system.

6. 11-4
• ATP-insensitive
Mutant
11-4
• Wild-Type
WT 11-4

7. • Using a forward
genetic
approach we
are identifying
the
chromosome on
which the gene
is present.

8. Seeds were harvested
from decedents of
different crosses
between the WT and
11-4 organisms. The
seeds were cleaned
and then sowed in a
MS Agar Media.

9. Earlier we
mentioned plant
responses to ATP
similar to those
found in
mammals.
One such
response is an
increase in
cytosolic calcium
concentrations
when ATP is
released into the
plant’s

10. Using a computer program I wrote, the data was
normalized and converted automatically into an Excel-
readable file, saving much time on what was previously a
largely manual process.

11. Wild Type
Here are some Mutant
examples of what the
normalized data
looks like when
graphed.
Non-Transgenic

12. Concentration (uM) Concentration (uM) Concentration (uM)
Plate 1 NoR HiR LoR NS Plate 7 NoR HiR LoR NS Plate 13 NoR HiR LoR no sig
1 = WT 3 8 0 1 = 6-2 0 5 3 4 1 = 4-1-1 0 4 8 0
2 = 11-4mut 10 2 0 2 = 6-3 1 0 1 10 2 = 4-1-2 0 4 8 0
3 = 1-1 7 3 2 3 = 6-4 0 3 6 3 3 = 4-1-3 8 1 3 0
4 = 1-2 2 5 5 4 = 7-1 2 3 3 4 4 = 4-1-4 2 5 5 0
5 = 1-3 4 3 5 5 = 7-2 4 2 3 3 5 = 4-2-1 8 0 4 0
6 = 1-4 5 2 5 6 = 8-1 0 1 7 4 6 = 4-2-2 1 5 6 0
7 = 1-5 4 5 3 7 = 8-2 2 4 4 2 7 = 4-2-3 8 1 3 0
8 = 1-6 7 4 1 8 = 8-3 0 3 5 4 8 = 4-2-4 3 1 8 0
Plate 2 NoR HiR LoR NS Plate 8 NoR HiR LoR NS Plate 14 NoR HiR LoR NS
1 = 8-4 3 2 3 4 1 = 4-3-1 4 3 3 2
2 = 9-2
3 = 9-4
0
2
0
5
2 10
3 2
2=2
3=4
1
8
6
3
5
1
0
0
This is a
4 = WQ 0 12 0 0 4 = 5-1-1 1 8 3 0
5=2
6=3
7=4
3
0
0
5
5
3
4
7
9
0
0
0
chart
Plate 3 NoR HiR LoR NS Plate 9
1=W
NoR HiR
0 0
LoR NS
0 12
8 = 5-2-1 0
Plate 15 NoR HiR
1 = 5-2-2
3 9
LoR NS
0
descriptiv
Key
Homo Mut
Homo Wild
2 = 11-4
3 = 2-1-1
4 = 2-1-2
0 0 0 12 2=3
3=4 e of the
Poss. Hetero 5 = 2-1-3
6 = 2-1-4
7 = 2-2-1
phenotypi
Plate 4 NoR HiR LoR NS 8 = 2-2-2
1.1 = WQ
1.2 = 11-4
2 = 2-1
0
3
2
4
2
3
2
1
5
0
0
2
Plate 10
1 = 2-2-3
2 = 2-2-4
NoR HiR
0
0
7
12
LoR NS
5 0
0 0
Plate 16 NoR HiR
1 = 3-2
2 = 3-7
8
0
0
0
LoR NS
1
0
3
12
c natures
3 = 2-2
4 = 2-3
5 = 2-4
1
1
2
7
7
2
1
2
6
3
2
2
3 = 2-3-1
4 = 2-3-2
5 = 2-3-3
0
0
0
12
12
9
0 0
0 0
3 0
3 = 4-2
4 = 12-2
5 = 12-1
2
2
9
0
3
1
3
1
2
7
6
0
of each
6 = 2-5
7 = 2-6
8 = 2-7
1
1
2
5
7
3
3
2
4
3
2
3
6 = 2-3-4
7 = 2-4-1
8 = 2-4-2
0
0
0
11
6
5
1 0
6 0
7 0
6 = 12-3
7 = 12-4
8 = 13-1
2
9
8
2
0
1
2
2
1
6
1
2
genetic
Plate 5 NoR HiR LoR NS Plate 11 NoR HiR LoR NS Plate 17 NoR HiR LoR NS
1 = 2-9
2 = 2-8
0
5
0
0
0
0
12
7
1 = 2-4-3
2 = 2-4-4
3
9
2
3
7 0
0 0
1 = 14-1
2=2
7
6
0
0
2
5
3
1
line.
3 = 2-10 3 2 1 6 3 = 3-1-1 10 1 1 0 3 = 15-1 0 0 6 6
4 = 2-11 3 2 0 7 4 = 2-5-1 9 1 2 0 4=2 9 1 0 2
5 = 2-12 0 11 1 0 5 = 3-1-2 12 0 0 0 5=3 3 2 4 3
6 = 2-13 8 0 2 2 6 = 3-1-3 6 0 6 0 6=4 6 2 1 3
7 = 2-14 1 9 3 0 7 = 3-1-4 7 0 5 0 7 = 16-1 6 0 1 5
8 = 2-15 5 2 1 4 8 = 3-2-1 10 0 2 0 8=2 5 1 0 6
Plate 6 NoR HiR LoR NS Plate 12 NoR HiR LoR NS Plate 18 NoR HiR LoR NS
1 = 2-16 3 1 6 2 1 = 3-2-2 11 0 0 1 1 = 16-6 2 3 2 5
2 = 2-17 0 6 2 4 2 = 3-2-3 8 1 3 0 2 = WQ 4 4 2 2
3 = 2-18 2 4 4 2 3 = 3-3-1 2 4 2 4
4 = 2-19 2 7 3 0 4 = 3-3-2 0 3 1 8
5 = 2-20 0 2 10 0 5 = 6-1 11 0 1 0
6 = 2-21 1 0 2 9 6 = 6-2 3 0 1 8
7 = 2-22 2 5 1 4 7 = 6-4 9 1 2 0
8 = 2-23 2 2 4 4 8 = 3-2-4 5 5 2 0

13. • The specimens that were
most likely homozygous-
mutants were selected
for genetic testing.
• PCR was used to amplify
specific areas of the
CONTROL
gene.
• A four-percent gel- Col Het Ler Col Het Col Het Het Ler Het Col Het
agarose was used in a Gel from nga172
gel electrophoresis
process to separate and
categorize each
specimen’s genotype.

14. These charts represent the gene and how many of our
specimens shared the genotype of the 11-4 organism.
CHROMOSOME NAME TEMP #Col #H+L #Het #Lan RECOMB CHI-SQUARE RATING Temp
nga63 55 0 #DIV/0! 0.007090101 48
ciw12 48 1 8 5 3 0.611111111 0.379589586 Great 52
SO392 55 0 #DIV/0! 0.007090101 55
CHROM 1 ciw1 48 0 #DIV/0! 0.007090101 Ind.
nga280 55 3 6 2 4 0.555555556 0.495056708 Great Rating
NF5|14 52 1 8 6 2 0.555555556 0.379589586 Great Ind. (Indescernable)
ATPase 55 2 7 1 6 0.722222222 0.848694963 Great Bad
nga1145 55 3 0 0 0 0 0.006169899 Bad Poor
CHROM 2 ciw3 48 0 0 0 0 #DIV/0! 0.002054719 Poor Ind. Good
nga1126 55 2 7 2 5 0.666666667 0.848694963 Poor Great
nga168 55 2 0 0 0 0 0.006981285 Bad
ng172 52 2 7 5 2 0.5 0.848694963 Great Gene?
nga126 52 0 0 0 0 #DIV/0! 0.002054719 Ind. LOW
CHROM 3 nga162 52 3 6 5 1 0.388888889 0.495056708 Good
ciw11 48 0 0 0 0 #DIV/0! 0.002054719 Ind.
ciw4 52 7 2 0 2 0.222222222 0.000200168 Good MED
nga6 52 3 4 0 4 0.571428571 0.191388523 Poor
ciw5 48 1 8 6 2 0.555555556 0.379589586 Great
nga8 52 6 2 2 0 0.125 0.001522054 Great HIGH
CHROM 4 ciw6 52 3 6 1 5 0.611111111 0.495056708 Great IND.
ciw7 48 0 #DIV/0! 0.007090101
nga1139 52 2 7 4 3 0.555555556 0.848694963 Good
nga1107 48 5 4 1 3 0.388888889 0.028163881 Good
nga225 52 8 1 1 0 0.055555556 7.41771E-06 Bad
nga249 52 4 5 3 2 0.388888889 0.151271016 Good
nga151 52 4 5 3 2 0.388888889 0.151271016 Great
CHROM 5 nga139 55 0 #DIV/0! 0.007090101
PHYC3 52 2 1 1 0 0.166666667 0.019956349 Bad
ciw9 52 3 5 1 4 0.5625 0.33015855 Great
ciw10 48 0 #DIV/0! 0.007090101

15. 1-4 2-13 2-20 3-2 12-4 13-1 14-1 15-2 12-1
nga63
ciw12 c l h h h h l l h
SO392
ciw1
nga280 c l h l h c l l c
NF5|14 c h h l h h h l h
ATPase c h c l l l l l l
nga1145 c l l l l c l c
ciw3
nga1126 c h l h l l l l c
nga168 c l l l l l c l h
ng172 c h c h h l h l h
nga126
nga162 c c c h h l h h h
ciw11
ciw4 c c c l c c c c l
nga6 c c l l c? c? l c l
ciw5 c h h l h h h l h
nga8 c c c c c c c c
ciw6 c l l h l l c l c
ciw7
nga1139
nga1107 c c l h c l c c l
nga225 c c c l c c c c c
nga249 c c l l h c h c h
nga151 c h l l h c h c c
nga139
PHYC3
ciw9 c c l? h c l l l l
ciw10

16. After analyzation of the
data, it was determined
that the gene of interest
was located close to the
nga8 region on the
fourth chromosome.

17. • Continue mapping with more specimens to further clarify
the data and ensure its accuracy.
• Continue mapping with different markers to get high-
resolution mapping results.
• Engineer a program to fully automate the normalization
and graphing process used in this experiment.
• Perform whole genome sequencing using BC1F3 back-
cross line to identify mutation point.

18. To Kiwamu
Tanaka,
Deanna
Lankford, Gary
Stacey, and the your time today, as well as
For
members of thefor these interesting and
Stacey Lab. educational past two months. I
have gained useful lab
experience, and been given a
unique insight into the world of
research.