BIO 120L lab8
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Simulating Methods to Estimate Population Size
Jordan Gum
3/27/23
1
Data
Activity 1
Data Table 1
Distanc
e
Transe
ct 1
Trial 1
Transe
ct 1 Trial 2
Transe
ct 2 Trial 1
Transe
ct 2
Trial 2
Transe
ct 3 Trial 1
Transe
ct 3 Trial 2
Transe
ct 4 Trial 1
Transe
ct 4 Trial 2
Transe
ct 5 Trial 1
Transe
ct 5 Trial 2
0 mm
blue
blue
blue
green
blue
green
red
blue
blue
Red
5 mm
Blue
green
Red
blue
red
blue
green
blue
blue
blue
10 mm
Red
red
blue
blue
blue
blue
blue
green
Red
blue
15 mm
blue
red
Green
blue
red
blue
red
blue
blue
blue
20 mm
red
blue
blue
blue
green
red
blue
red
blue
Red
25 mm
Green
Blue
blue
red
blue
blue
blue
green
blue
Gree
n
30 mm
green
Blue
green
blue
blue
red
blue
green
red
Blue
35 mm
red
red
blue
green
blue
blue
blue
blue
Green
Blue
40 mm
blue
blue
red
red
green
blue
red
blue
blue
Red
45 mm
red
red
blue
red
blue
red
blue
blue
red
Red
50 mm
blue
blue
blue
red
red
red
red
red
blue
Blue
55 mm
green
blue
blue
blue
red
green
red
blue
red
Red
60 mm
green
red
blue
green
green
blue
red
red
blue
Blue
65 mm
blue
red
red
blue
blue
red
red
green
red
Blue
70 mm
red
green
red
blue
blue
blue
blue
blue
green
Blue
75 mm
blue
blue
blue
blue
green
blue
blue
blue
red
Red
80 mm
blue
red
blue
red
blue
blue
red
blue
blue
Blue
85 mm
red
blue
red
red
red
red
blue
blue
blue
Red
90 mm
blue
blue
blue
green
red
green
blue
red
blue
Blue
95 mm
red
blue
green
blue
green
red
blue
green
green
Blue
100 mm
blue
green
blue
blue
blue
blue
red
blue
red
gree
n
© 2016 Carolina Biological Supply Company
2
Data Table 2
Color
Trial 1
Total
Trial 1
Percent
Coverage
Trial 2
Total
Trial 2
Percent
Coverage
Blue
55
53%
57
55%
Red
34
32%
31
29%
Green
16
15%
17
16%
Total Number of Squares Counted
105
105
105
105
1.
In terms of percent coverage rank the colors on the Population Sheet from
highest to lowest.
Blue, Red, and Green would be the order of the hues. Red makes up 20% of the area (mine was 29-32%), Green makes up 15% of the area (close to my range of 15–16%), and Blue makes up 65% of the actual region.
2.
How would the data change if 10 transects were run during this activity?
This will differ; however, as more samples are taken and transects are added, there should be less variation within and between groups.
3.
In addition to percent cover, what other information could be determined about each population using the transect data that was collected?
The size and density of the populace (the number of people per unit of area or volume) can be calculated.
4.
List three situations or types of organisms for which this type of population study is appropriate.
This will differ. There are a variety of vegetation kinds in the wild,
forest tree cover, and sessile (non-moving) aquatic creatures as potential candidates. For organisms that travel very little or not at all, transects are the best option.
© 2016 Carolina Biological Supply Company
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7.
8.
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a. What are the larger round cells (general name)?
Header
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446 PM
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9
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1696299405-SBI4U Test...
&
Preview File Edit View Go Tools Window Help
4
Apoft being educach pr
er her s
me
Dining
may
waters and
the day of the act of DCA I har ful
4
Part-Thinking and Ingairy (15)
Camp Bukaryti gone uplates to Prakaryotic game regulation. What the
3. modderpoed for Deplication What they called How are they dont
5
P
5. Dedimxplain the process of transcription. Add
4 Frock point todemontherf of genc
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انادا در در درست
1696299405-SBI4U Test 3.pdf
Page 6 of 10
days of this course. Please remember the Dropboxes will close
22
280
JAN
22
4. Outline the steps of Frederick Griffith's experiment on mice to demonstrate the transfer of genetic
material using Streptococcus pneumoniae.
tv
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U₁
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9
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80
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Efficacy and safety of herbal medicine (Lianhuaqingwen) for treating COVID 19
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Clal - Bap 23
Hindill 29
Aatll- Iral 4284
B 4209
B 4205
Ecok 185
Spl 4168
Batl- Nel 229
Al 404
an 35
Sgrl 400
hanll 471
Scal 3844
Ball 485
378
Prul 3733-
Kee 22
P 300
tr 3002
Sall - let-indl 651
P 712
At 3537
Bal 3433
Tag 09
hel
Neel 92
T 3420
PBR322
4361 bp
BePI 1045
Bep - a 103
PEMI 1315
PENE 1364
Sl 1360
Awal - 1425
u 1438
Msd 1444
t 14
Pyu 140
Aul 3000
Aa 2884
Bell 2777
rep
ch 22
Dell 2575
Bag 1650
Bp 1664
Bal 16
Pal - M 2473
204
farl 2351
- Sapt 20
Ndet 226
2291
BI- Nd 2244/
Ba 2225
Xnel 2009
Pull 2064
Ben 2122
Dell 2162
Tih- PI 217
Figure 1
If your gene of interest was inserted at the Sphl restriction site of the plasmid illustrated in
Figure 1, describe the screening process to select the positive recombinants.
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13-14. A set of Hfr strains transfer markers in the following order.
Strain
Transfer order
JIAF G
1
ED BGFA
IJH CE D
CED B
13. Which of the following gene orders is correct?
A: AFGEDBCHIJ
B: ECHIJAFGBD
C: E D B G FACH JI
D: E D B G FACHIJ
E: A F G BD ECHJI
14. Which Hfr strains have F factors inserted in the same orientation?
A: 1 and 3 oriented the same way, 2 and 4 both oriented the other way
B: 1 and 2 oriented the same way, 3 and 4 both oriented the other way
C: 1 and 4 oriented the same way, 2 and 3 both oriented the other way
D: 1 oriented one way, 2, 3 and 4 all oriented the other way
E: all are oriented the same way
2
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FOR ACADEMIC PURPOSES ONLY - NOT FOR CLINICAL USE
Test Details
Gene
Genotype
Phenotype
Alleles Tested
CYP2C19
*2/*7
Poor Metabolizer
*2, *3, *4, *48, *5, *6, *7, *8, *9, *17
CYP2D6
*1/4
Intermediate Metabolizer
*2, *3, *4, *4M, "6, *7, *8, *9, *10, *12, *14A, *148, *17, *29,
*35, *41
CYP3AS
*3/*3
Poor Metabolizer
*1D, *2, *3, *38, *3C, "6, *7, "8, *9
CYP3A4
*1/*1
Normal Metabolizer
*18, *2, *3, *12, *17, *22
VKORCI
-1639G>A G/A
Intermediate Warfarin Sensitivity
-1639G>A
*2, *3
*2, *3, *4, *5, "6, *11
CYP4F2
*1/3
Decreased Function
CYP2C9
*1/*1
Normal Metabolizer
CYP2B6
*1/*1
Normal Metabolizer
*6, *9
Normal Metabolizer - Higher
Inducibility
CYPIA2
*1F/"1F
*1C, *1D, *1F, *1K, *1L, *1V, *1W
SLCOIB1
521T>CT/T
Normal Function
521T>C, 388A>G
CFTR
F508del/RS53X
Negative
Numerous
DPYD
*1/*1
Normal Metabolizer
Numerous
ТРМТ
*1/*3A
*1/1
Intermediate Metabolizer
Normal Metabolizer
*2, *3A, *38, *3C, *4
*2, *3, *4, *5, "6, *7, *8, *9
NUDT15
UGTIAI
*1/*36
Normal Metabolizer
*6, *27, *28, *36, *37,…
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Sample Number
Grid Coordinate
Letter
Grid Coordinate
Number
Number of dandelions in quadrat
Sample Number
Grid coordinate
Letter
Grid Coordinate
Number
Number of Dandelions
in quadrat
1
H
12
0
11
D
13
1
2
A
14
0
12
G
1
0
3
H
1
0
13
C
10
2
4
J
14
1
14
A
6
1
5
C
2
1
15
I
7
2
6
I
3
2
16
E
14
1
7
D
11
3
17
C
5
2
8
F
9
3
18
D
3
3
9
H
8
2
19
G
8
2
10
F
7
2
20
G
2
2
Find the average number of dadelions.
Find the percent error
Find the population estimate.
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AaBbCcDc AaBbC
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Activity 1.
3.
Based on the following data which of the following would be considered:
a. Recessive?
b. Dominant?
C. A weak loss-of-function mutation?
d. A null mutation?
e. A haploinsufficiency ?
Data - Gene "B" is required for formation of body segments. A wild type fly had 3 body
segments.
BB (Homozygous
dominant)
Wild Type
Bb (Heterozygous)
bb (Homozygous
recessive)
Lacks all body
segments
1 body segment
Mutant 1
Wild Type
Mutant 2
Wild Type
2 body segments
Mutant 3
Wild Type
Wild Type
2 body segments
Activity 4
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1. List 4 methods general methods for experimentally inhibiting the action of specific genes.
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A BI-201 Syllabus Virtual Spring 2021.pdf (page 1 of 19)
A BioLab Fly_Group_F.pdf (page 4 of 4)
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v BioLab Fly_Group_F.pdf
HIST111
Mutation: Star eye shape
signment1
P generation Phenotypes:
Normal
Star
2
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male
DOCX
W1 - Puffins
Fi generation
Phenotvne
Females
Males
Total
Ratio
Normal
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021-04..3.03 AM
F, XF, Phenotypes:
Normal
female X
Normal
male
DOCX
F2 generation
HIST111
Phenotvoe
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HIST111 2
Atomic
This mutation is inherited as:
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dominant autosomal (da)
recessive autosomal (ra)
recessive sex-linked (rs)
DOCX
History
homework4
APR
29
X
HENE
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Rh-hr
Kell
Duffy
Xg Results
D C E с
Kp Kp Js Js Fy Fy
Xg AHG 0.2
M
DTT
+
+
+
0 +
+ 2+ 2+
+ 0
0 0
0
+
0
+
0
0
0
0
+ Neg Neg
0
0
+ 0
+
+
0
+
0
0
+
+2+ Neg
+2+ 2+
+
+
+
+
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+
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+
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+
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+
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+
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+
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0 Neg Neg
ID4
0 +
+
0
0
0
+
0
+ 0
+
+2+ 2+
Neg Neg
ID5
0 0
+
0
0
+
0
+
+
0
+ 0
+
+
ID6
0 0
+
0
+
+
0
0
0
0
+
+
+
0
+
0 2+ Neg
ID7
0 0
+
0
0
+
0
0
0
+
0
0 0
0
+
0
0
+
+ Neg Neg
ID8
+ 0
+ +
0
0 +
0
+
0 0
+
0
0
+ 0
+
+
0
+
+ Neg Neg
ID9
0 0
+
0
0
0
+
0
0
+
+
0
0
0
+ 0
+
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0 Neg Neg
ID10 000
+
0
0
+
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+ 0
0
0
+
0
+
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+
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ID11
000
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0
+
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+Neg Neg
AC
Neg
AC, auto control; SC, screening cells; ID, identification panel; Neg. Negative reaction
Note: SC1 to SC3 are screening cells, ID1 to ID11 are panel cells. All screening and panel cells have negative results with the IS phase and Thermo
Phase.
1. What is the result of the Antibody Screening test?
2. Which of the following panel cells contain only the Js" antigen?
3.…
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A clinical trial was carried out to test whether a new treatment affects the recovery rate of patients suffering from a debilitating disease. The null hypothesis “H0: The treatment has no effect” was rejected with a P-value of 0.04. The researchers used a significance level of a = 0.05. State whether each of the following conclusions is correct. If not, explain why:
a)The treatment has only a small effect ()
b)The treatment has some effect ()
c)The probability of committing a Type I error is 0.04 ()
d)The probability of committing a Type II error is 0.04 ()
e)The null hypothesis would not have been rejected had the significance level been set to a = 0.01 instead of 0.05. ()
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- FOR ACADEMIC PURPOSES ONLY - NOT FOR CLINICAL USE Test Details Gene Genotype Phenotype Alleles Tested CYP2C19 *2/*7 Poor Metabolizer *2, *3, *4, *48, *5, *6, *7, *8, *9, *17 CYP2D6 *1/4 Intermediate Metabolizer *2, *3, *4, *4M, "6, *7, *8, *9, *10, *12, *14A, *148, *17, *29, *35, *41 CYP3AS *3/*3 Poor Metabolizer *1D, *2, *3, *38, *3C, "6, *7, "8, *9 CYP3A4 *1/*1 Normal Metabolizer *18, *2, *3, *12, *17, *22 VKORCI -1639G>A G/A Intermediate Warfarin Sensitivity -1639G>A *2, *3 *2, *3, *4, *5, "6, *11 CYP4F2 *1/3 Decreased Function CYP2C9 *1/*1 Normal Metabolizer CYP2B6 *1/*1 Normal Metabolizer *6, *9 Normal Metabolizer - Higher Inducibility CYPIA2 *1F/"1F *1C, *1D, *1F, *1K, *1L, *1V, *1W SLCOIB1 521T>CT/T Normal Function 521T>C, 388A>G CFTR F508del/RS53X Negative Numerous DPYD *1/*1 Normal Metabolizer Numerous ТРМТ *1/*3A *1/1 Intermediate Metabolizer Normal Metabolizer *2, *3A, *38, *3C, *4 *2, *3, *4, *5, "6, *7, *8, *9 NUDT15 UGTIAI *1/*36 Normal Metabolizer *6, *27, *28, *36, *37,…arrow_forward4a. what is concordance rate 4b. what is discordance rate 9c. what do the results suggest? alcoholism is genetic? alcohlosim is enviromental? no indication of either 4d. what if studies were compared to study of dizygotic twins with a concordance rate of 30%z what do results suggest? alcolism is genetic alcolism is enviromental? they are equal? no indication?arrow_forwardSample Number Grid Coordinate Letter Grid Coordinate Number Number of dandelions in quadrat Sample Number Grid coordinate Letter Grid Coordinate Number Number of Dandelions in quadrat 1 H 12 0 11 D 13 1 2 A 14 0 12 G 1 0 3 H 1 0 13 C 10 2 4 J 14 1 14 A 6 1 5 C 2 1 15 I 7 2 6 I 3 2 16 E 14 1 7 D 11 3 17 C 5 2 8 F 9 3 18 D 3 3 9 H 8 2 19 G 8 2 10 F 7 2 20 G 2 2 Find the average number of dadelions. Find the percent error Find the population estimate.arrow_forward
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