Required information In Drosophila- the autosomal recessive prand cn m.pdf
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Required information In Drosophila, the autosomal recessive prand cn mutations cause brown and bright-red eyes, respectively (wild-type flies have brick-red eyes). Flies who are homozygous recessive at both prand c n have orange eyes. A female who has wildtype eyes is crossed to an orange-eyed male. Their progeny have the following distribution of eye colors: he mother of these progeny resulted from a cross between two flies from true-breeding lines. What are the genotypes of these two lines + Cr + / r en' and pr en/ pren px en' i pron and prent or on pren'i prent and p x + onl per cn t orl pren and prent/ pron More then coe of these could be trie. p r + cn + / p r + cn + and p rc n / p rc n p r + cn + / pr c n and p rc n / pr c n p r Cn + / p r Cn + and p r + cn / p r + cn p r + c n / p rc n and p rc n + / p rc n More than one of these could be true..
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A three point cross was performed using fruit flies. The three loci .pdf
A three point cross was performed using fruit flies. The three loci are physically linked on the
same autosome. The d allele, in the homozygous state, results in the recessive ‘double dorsal’
phenotype. Similarly, the r allele results in the recessive ‘round ears’ phenotype and the e allele
in the recessive ‘endless eye’ phenotype. In contrast, the D, R, and E alleles encode for the
dominant wild type phenotypes at each of these loci.
True breeding (homozygous) flies were crossed to produce F1 offspring. Due to poor book-
keeping, the genotypes of the parent flies were not recorded – but one parent fly had one mutant
phenotype, while the other parent had the other two mutant phenotypes.
PhenotypeNumber of F2 Flies
EarsDorsalEyes
1) Round ears Double Dorsal Endless Eyes 230
2) Wild-type Wild-type Wild-type 260
3) Round ears Wild-type Wild-type 4,390
4) Wild-type Double Dorsal Endless Eyes 4,100
5) Round ears Double Dorsal Wild-type 15
6) Wild-type Wild-type Endless Eyes 15
7) Round ears Wild-type Endless Eyes 510
8) Wild-type Double Dorsal Wild-type 480
10,000 total
a) What is the phenotype of the F1 flies?
b) What is the genotype of the F1 flies? Your answer should indicate the phase of linkage.
c) What are the genotypes and phenotypes of the original parent flies (previously not recorded)?
Your answer should indicate the phase of linkage.
d) Calculate the map distance in cM between each of the three loci and determine the gene order.
Show all work for full credit. (6 points)
Solution
a. The phenotype of the F1 flies is wild-type heterozygous at all locus.
b. The genotype of F1 flies
r D E/ R d e
c. The genotypes of parent flies: r D E and R d e
The phenotypes of parent flies: Round ears, Wild-type, Wild-type and Wild-type, Double Dorsal,
Endless Eyes
d.
Gene order is R------D------E
The distance between R and D
510+480+15+15 / 10,000 * 100 = 10.2cM
The distance between D and E
230+260+15+15 / 10,000 * 100 = 5.2cM.
In fruit flies, the brown gene, which controls eye color, is located .pdf
In fruit flies, the brown gene, which controls eye color, is located on an autosome. Brown
mutants (bw) have brown eyes, while wild-type flies (bw^+) have red eyes. A gene on the X
chromosome, miniature, controls wing size. Miniature wings are caused by the m allele, which is
recessive to the allele for normal wings (m^+). True-breeding miniature-winged, red-eyed
females are crossed to true-breeding normal-winged, brown-eyed males. Predict the phenotypes
(including their ratios of the F1 flies. If the F1 flies are intercrossed with each other, what
phenotypes will appear in the F2 and in what proportions?
Solution
1.
Phenotype
Normal wings Red Eyes 4418
Miniature wings Brown eyes 4330
Normal wings Brown eyes 610
Miniature wings Red eyes 590
2. The F1 was self-crossed to give rise to the F2 generation. Of these offspring, 3/4 had brown
eyes while 1/4 had Red eyes. Phenotypic ratio is 3:1.
The phylogeny in this figure shows the evolution of visual opsins in.pdf
The phylogeny in this figure shows the evolution of visual opsins in tetrapods. Each colored
triangle represents a different opsin. Which of the following accurately describes the opsins in
these species? The phylogeny in this figure shows the evolution of visual opsins in tetrapods.
Each colored triangle represents a different opsin. Which of the following accurately describes
the opsins in these species? A AAA AAAA A A AA AAA AAA Squamate reptiles New World
A opsin gain A opsin loss Tetrapod Click to view larger im O The green opsin is a
synapomorphy in rodents and New World primates, and the light blue opsin is a
symplesiomorphy in baboons and humans. O b. The light blue opsin is a synapomorphy in
squamate reptiles and birds, and the green opsin is an ancestral trait. O c The purple opsl in is an
analogous trait, and the green opsin is a homologous trait. O d The green opsin is a
synapomorphy in baboons and humans, and the light blue opsin is symplesiomorphy in squamate
reptiles and birds.
Solution
synapomorphy means a character that is present in an ancestral species and then it is shared by
one or more descendants, symplesiomorphy means that it is a shared primitive trait.
since we can see that the green opsin is first present in the ancestor the tetrapod and then gets
transferred to only baboons and humans it is derived, also we see that light blue opsin is
transferred to the reptiles and the birds from the tetrapod ancestor.
therefore the correct option should be:
the green opsin is a synapomorphy in baboons and humans, and the light blue opsin is
symplesiomorphy in squamate reptiles and birds..
InfoRefsDataWorldMapMain.docx
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Premise
DNA Evidence is routinely used as a means to determine the identity of a suspect that has left DNA at a crime scene.
If a suspect does not match the evidence left at a crime scene, they can be excluded with near certainty.
If the suspect does share his/her genotype with the evidence, he can not be excluded as the perpetrator.
The number of loci genotyped and the allele frequencies represented determine the likelihood of a false positive match.
FREDONIA
Damn Footprints!!!
Backstory
A crime has been committed in the new Science Center. Over the winter break a trespasser broke into the science center and walked across the freshly poured floors in their dirty boots. As a result the floors in the second floor hallway has been ruined and must be replaced. University police arrived on the scene in time to see a single person running from the worksite. While the perpetrator was not apprehended, they did loose their hat on the fence as they made their escape. Several hairs were found in the hat with the follicles intact. The police have provided us with DNA extracted from these follicles in hopes that we can identify the perpetrator and bring them to justice. It is up to you to solve this hairy crime and identify the person responsible for delaying the completion of the Biology department's new home.
Premise
Failure to exclude a suspect is not the same as proving that he/she is the perpetrator.
Confidence is assessed by considering several loci and considering the allele frequency at each locus.
By using neutral polymorphisms (no selective pressure), estimates can be made about the expected frequency of the indicated alleles randomly occurring in the population.
Such loci conform to Hardy-Weinberg equilibrium.
Population Genetics
Population: A group that lives in the same geographic area and can successfully reproduce.
Gene Pool: Genes present in a population.
Allele Frequency: For a specific gene, the relative frequency of each allele. The sum of all allele frequencies must = 1.
Calculation of allelic frequencies in gametes
Hardy-Weinberg Equilibrium
p + q = 1
p = allele frequency of one allele
q = allele frequency of a second allele
p2 + 2pq + q2 = 1
p2 and q2 genotype frequencies for each homozygote
2pq genotype frequency for heterozygotes
All of the allele frequencies together equal 1
All of the genotype frequencies
together equal 1
Genotype Frequencies In Progeny
Paternal Gametes
A(p) a(q)
Maternal A(p) AA(p2) Aa(pq)
Gametes a(q) Aa(pq) aa(q2)
genotype: AA Aa aa
frequency: p2 2pq q2
Or Mathematically: (p+q)(p+q) = p2 + 2pq + q2 = 1
A Punnett square showing the ratio p2:2pq:q2
Implications of the Hardy-Weinberg Principle
Allelic frequencies remain constant from generation to generation
p + q = 1
frequencies of AA, Aa, and aa genotypes
are p2, .
Problems2. This is a map for a diploid plantR--------35-------.pdf
Problems:
2. This is a map for a diploid plant:
R--------35--------E--------17---------F
One parental plant is R e f/r E F. It is mated with a homozygous recessive plant: r e f/r e f. Give
the proportion of the offspring assuming there is both no interference, and complete interference.
3. In a newly discovered insect, the three genes R, A, and P located on the same chromosome
(R= red body, A= spotted coloring, P=purple eyes, r=white body, a=solid coloring, p=white
eyes). A heterozygous parent(R A P/r a p) is mated with a homozygous recessive parent (r a p/ r
a p).
The following offspring are shown below:
(210) dominant for all traits
(16) red body, solid coloring, purple eyes
(70) red body, spotted coloring, white eyes
(50) red body, solid coloring, white eyes
(200) recessive for all traits
(55) white body, spotted coloring, purple eyes
(13) white body, spotted coloring, white eyes
(72) white body, solid coloring, purple eyes
Perform a three-point linkage to determine the gene order and relative distance between the
genes. How did the distance between the 2 genes affect the frequency of crossing-over events?
4. Here are four recombinant sequences that came from four Hfr matings during an interrupted
mating experiment. Determine the order and relative distance by setting A to 0/100 minutes.
Hfr1) A T1 H T2
8 26 33 53
Hfr2) E1 T1 E2 E3
4 12 24 54
Hfr3) T2 S T3 A
7 37 52 62
Hfr4) T3 C E1 E3
2 17 22 72
5. In a u-tube experiment, a researcher placed a kanamycin resistant strain of E. coli on one side
of a membrane impervious to the cells, and a wild-type strain on the other side. After a day, he
then plated a sample from both sides of the tube on plates containing kanamycin and observed
growth on both plates. He then repeated this experiment, but this time he added DNase to the
tube and observed the same results when he plated both cell types. What happened inside the
tube?
6. Pretend the following is a chromosome that underwent some changes. What were they? (the
dot is the centromere).
A B C D E F . G H I J A B C G . F E H I J
a. pericentric inversion, deletion
b. paracentric inversion, deletion
c. deletion, translocation
d. Robertsonian translocation,
pericntric inversion
7. Given:
Type of mutation Cistron A or B
1. point A
2. point A
3. point B
4. point B
5. deletion A
Results from 5 crosses:
1x2 (plaque) 2x3 (plaque) 3x4 (plaque) 4x5 (plaque)
1x3 (plaque) 2x4 (plaque) 3x5 (plaque)
1x4 (plaque) 2x5 (no plaque)
1x5 (plaque)
8. The following nucleotide sequence is found in a piece of DNA
5’-TTAACA-3’
3’-AATTGT-5’
if the cytosince undergoes a tautameric shift for one replication cycle draw two round of
replication and explain what type of mutation occurred (Transition or Transversion)?
Bonus:
Explain the relationship between mosaicism and x-inactivation using an example.
9. In a diploid fly, the three loci R E and D are linked as follows
R--------------22cm------------E----------16cm-------------D
One fly is available to you and its genotype is R.
Recommended
A three point cross was performed using fruit flies. The three loci .pdf
A three point cross was performed using fruit flies. The three loci are physically linked on the
same autosome. The d allele, in the homozygous state, results in the recessive ‘double dorsal’
phenotype. Similarly, the r allele results in the recessive ‘round ears’ phenotype and the e allele
in the recessive ‘endless eye’ phenotype. In contrast, the D, R, and E alleles encode for the
dominant wild type phenotypes at each of these loci.
True breeding (homozygous) flies were crossed to produce F1 offspring. Due to poor book-
keeping, the genotypes of the parent flies were not recorded – but one parent fly had one mutant
phenotype, while the other parent had the other two mutant phenotypes.
PhenotypeNumber of F2 Flies
EarsDorsalEyes
1) Round ears Double Dorsal Endless Eyes 230
2) Wild-type Wild-type Wild-type 260
3) Round ears Wild-type Wild-type 4,390
4) Wild-type Double Dorsal Endless Eyes 4,100
5) Round ears Double Dorsal Wild-type 15
6) Wild-type Wild-type Endless Eyes 15
7) Round ears Wild-type Endless Eyes 510
8) Wild-type Double Dorsal Wild-type 480
10,000 total
a) What is the phenotype of the F1 flies?
b) What is the genotype of the F1 flies? Your answer should indicate the phase of linkage.
c) What are the genotypes and phenotypes of the original parent flies (previously not recorded)?
Your answer should indicate the phase of linkage.
d) Calculate the map distance in cM between each of the three loci and determine the gene order.
Show all work for full credit. (6 points)
Solution
a. The phenotype of the F1 flies is wild-type heterozygous at all locus.
b. The genotype of F1 flies
r D E/ R d e
c. The genotypes of parent flies: r D E and R d e
The phenotypes of parent flies: Round ears, Wild-type, Wild-type and Wild-type, Double Dorsal,
Endless Eyes
d.
Gene order is R------D------E
The distance between R and D
510+480+15+15 / 10,000 * 100 = 10.2cM
The distance between D and E
230+260+15+15 / 10,000 * 100 = 5.2cM.
In fruit flies, the brown gene, which controls eye color, is located .pdf
In fruit flies, the brown gene, which controls eye color, is located on an autosome. Brown
mutants (bw) have brown eyes, while wild-type flies (bw^+) have red eyes. A gene on the X
chromosome, miniature, controls wing size. Miniature wings are caused by the m allele, which is
recessive to the allele for normal wings (m^+). True-breeding miniature-winged, red-eyed
females are crossed to true-breeding normal-winged, brown-eyed males. Predict the phenotypes
(including their ratios of the F1 flies. If the F1 flies are intercrossed with each other, what
phenotypes will appear in the F2 and in what proportions?
Solution
1.
Phenotype
Normal wings Red Eyes 4418
Miniature wings Brown eyes 4330
Normal wings Brown eyes 610
Miniature wings Red eyes 590
2. The F1 was self-crossed to give rise to the F2 generation. Of these offspring, 3/4 had brown
eyes while 1/4 had Red eyes. Phenotypic ratio is 3:1.
The phylogeny in this figure shows the evolution of visual opsins in.pdf
The phylogeny in this figure shows the evolution of visual opsins in tetrapods. Each colored
triangle represents a different opsin. Which of the following accurately describes the opsins in
these species? The phylogeny in this figure shows the evolution of visual opsins in tetrapods.
Each colored triangle represents a different opsin. Which of the following accurately describes
the opsins in these species? A AAA AAAA A A AA AAA AAA Squamate reptiles New World
A opsin gain A opsin loss Tetrapod Click to view larger im O The green opsin is a
synapomorphy in rodents and New World primates, and the light blue opsin is a
symplesiomorphy in baboons and humans. O b. The light blue opsin is a synapomorphy in
squamate reptiles and birds, and the green opsin is an ancestral trait. O c The purple opsl in is an
analogous trait, and the green opsin is a homologous trait. O d The green opsin is a
synapomorphy in baboons and humans, and the light blue opsin is symplesiomorphy in squamate
reptiles and birds.
Solution
synapomorphy means a character that is present in an ancestral species and then it is shared by
one or more descendants, symplesiomorphy means that it is a shared primitive trait.
since we can see that the green opsin is first present in the ancestor the tetrapod and then gets
transferred to only baboons and humans it is derived, also we see that light blue opsin is
transferred to the reptiles and the birds from the tetrapod ancestor.
therefore the correct option should be:
the green opsin is a synapomorphy in baboons and humans, and the light blue opsin is
symplesiomorphy in squamate reptiles and birds..
InfoRefsDataWorldMapMain.docx
In
fo
R
ef
s
D
at
a
W
or
ld
M
ap
M
ai
n
Premise
DNA Evidence is routinely used as a means to determine the identity of a suspect that has left DNA at a crime scene.
If a suspect does not match the evidence left at a crime scene, they can be excluded with near certainty.
If the suspect does share his/her genotype with the evidence, he can not be excluded as the perpetrator.
The number of loci genotyped and the allele frequencies represented determine the likelihood of a false positive match.
FREDONIA
Damn Footprints!!!
Backstory
A crime has been committed in the new Science Center. Over the winter break a trespasser broke into the science center and walked across the freshly poured floors in their dirty boots. As a result the floors in the second floor hallway has been ruined and must be replaced. University police arrived on the scene in time to see a single person running from the worksite. While the perpetrator was not apprehended, they did loose their hat on the fence as they made their escape. Several hairs were found in the hat with the follicles intact. The police have provided us with DNA extracted from these follicles in hopes that we can identify the perpetrator and bring them to justice. It is up to you to solve this hairy crime and identify the person responsible for delaying the completion of the Biology department's new home.
Premise
Failure to exclude a suspect is not the same as proving that he/she is the perpetrator.
Confidence is assessed by considering several loci and considering the allele frequency at each locus.
By using neutral polymorphisms (no selective pressure), estimates can be made about the expected frequency of the indicated alleles randomly occurring in the population.
Such loci conform to Hardy-Weinberg equilibrium.
Population Genetics
Population: A group that lives in the same geographic area and can successfully reproduce.
Gene Pool: Genes present in a population.
Allele Frequency: For a specific gene, the relative frequency of each allele. The sum of all allele frequencies must = 1.
Calculation of allelic frequencies in gametes
Hardy-Weinberg Equilibrium
p + q = 1
p = allele frequency of one allele
q = allele frequency of a second allele
p2 + 2pq + q2 = 1
p2 and q2 genotype frequencies for each homozygote
2pq genotype frequency for heterozygotes
All of the allele frequencies together equal 1
All of the genotype frequencies
together equal 1
Genotype Frequencies In Progeny
Paternal Gametes
A(p) a(q)
Maternal A(p) AA(p2) Aa(pq)
Gametes a(q) Aa(pq) aa(q2)
genotype: AA Aa aa
frequency: p2 2pq q2
Or Mathematically: (p+q)(p+q) = p2 + 2pq + q2 = 1
A Punnett square showing the ratio p2:2pq:q2
Implications of the Hardy-Weinberg Principle
Allelic frequencies remain constant from generation to generation
p + q = 1
frequencies of AA, Aa, and aa genotypes
are p2, .
Problems2. This is a map for a diploid plantR--------35-------.pdf
Problems:
2. This is a map for a diploid plant:
R--------35--------E--------17---------F
One parental plant is R e f/r E F. It is mated with a homozygous recessive plant: r e f/r e f. Give
the proportion of the offspring assuming there is both no interference, and complete interference.
3. In a newly discovered insect, the three genes R, A, and P located on the same chromosome
(R= red body, A= spotted coloring, P=purple eyes, r=white body, a=solid coloring, p=white
eyes). A heterozygous parent(R A P/r a p) is mated with a homozygous recessive parent (r a p/ r
a p).
The following offspring are shown below:
(210) dominant for all traits
(16) red body, solid coloring, purple eyes
(70) red body, spotted coloring, white eyes
(50) red body, solid coloring, white eyes
(200) recessive for all traits
(55) white body, spotted coloring, purple eyes
(13) white body, spotted coloring, white eyes
(72) white body, solid coloring, purple eyes
Perform a three-point linkage to determine the gene order and relative distance between the
genes. How did the distance between the 2 genes affect the frequency of crossing-over events?
4. Here are four recombinant sequences that came from four Hfr matings during an interrupted
mating experiment. Determine the order and relative distance by setting A to 0/100 minutes.
Hfr1) A T1 H T2
8 26 33 53
Hfr2) E1 T1 E2 E3
4 12 24 54
Hfr3) T2 S T3 A
7 37 52 62
Hfr4) T3 C E1 E3
2 17 22 72
5. In a u-tube experiment, a researcher placed a kanamycin resistant strain of E. coli on one side
of a membrane impervious to the cells, and a wild-type strain on the other side. After a day, he
then plated a sample from both sides of the tube on plates containing kanamycin and observed
growth on both plates. He then repeated this experiment, but this time he added DNase to the
tube and observed the same results when he plated both cell types. What happened inside the
tube?
6. Pretend the following is a chromosome that underwent some changes. What were they? (the
dot is the centromere).
A B C D E F . G H I J A B C G . F E H I J
a. pericentric inversion, deletion
b. paracentric inversion, deletion
c. deletion, translocation
d. Robertsonian translocation,
pericntric inversion
7. Given:
Type of mutation Cistron A or B
1. point A
2. point A
3. point B
4. point B
5. deletion A
Results from 5 crosses:
1x2 (plaque) 2x3 (plaque) 3x4 (plaque) 4x5 (plaque)
1x3 (plaque) 2x4 (plaque) 3x5 (plaque)
1x4 (plaque) 2x5 (no plaque)
1x5 (plaque)
8. The following nucleotide sequence is found in a piece of DNA
5’-TTAACA-3’
3’-AATTGT-5’
if the cytosince undergoes a tautameric shift for one replication cycle draw two round of
replication and explain what type of mutation occurred (Transition or Transversion)?
Bonus:
Explain the relationship between mosaicism and x-inactivation using an example.
9. In a diploid fly, the three loci R E and D are linked as follows
R--------------22cm------------E----------16cm-------------D
One fly is available to you and its genotype is R.
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Issued 9,400 shares of common stock at $5.30 per share.
Issued 20,400 shares of common stock at $10.10 per share.
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Paid dividends of $53,000.
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insur- The fame of his Merrymen was spreading, and new rection against the High Sheriff of
Nottingham recruits were pouring in from every corner of that Robin Hood took a walk in
Sherwood For- England. As the band grew larger, their small bivouac est. As he walked he
pondered the progress of the became a major encampment Between raids the men campaign, the
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moves, and the options that confronted him. Was in decline, and discipline was becoming harder
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the Sheriff and his administration. However, alone The growing band was also beginning to
exceed Robin Hood could do little. He therefore sought al- the food capacity of the forest. Game
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be obtained from outlying Later he welcomed all who came, asking few ques- villages. The cost
of buying food was beginning to tions and demanding only a willingness to serve. drain the
bund's financial reserves at the very moment Strength, he believed, lay in numbers. He spent the
first year forging the group into a those with the most to lose, were now giving the fordisciplined
band, united in enmity against the Sheriff est a wide berth. This was costly and inconvenient to
and willing to live outside the law. The band's organi- them, but it was preferable to having all
their goods zation was simple. Robin ruled supreme, making all confiscated. important decisions.
He delegated specific tasks to Robin believed that the time had come for the his lieutenants. Will
Scarlett was in charge of intel- Merrymen to change their policy of outright confisligence and
scouting. His main job was to shadow the cation of goods to one of a fixed transit tax. His
lieuSheriff and his men, always alert to their next mowe. tenants strongly resisted this idea. They
were proud He also collected information on the travel plans of of the Merrymen's famous motto:
"Rob the rich and rich merchants and tax collectors. Little John kept dis. give to the pooc." -The
farmers and the townspeocipline among the men and saw to it that their archery ple," they
argued, "are our most important allies. was at the high peak that their profession demanded. How
can we tax them, and still bope for their help in Searlock took care of the finances, converting
loot to our fight against the Sherif?? cash, paying shares of the take, and finding suitable Robin
wondered how long the Merrymen could hiding places for the sarplus. Finally, Much the keep to
the ways and methods of their early days. Miller's son had the difficult task of provisioning the
The Sheriff was growing stronger and becoming.
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prenatal retardation or disruption of cell division, leading to malformation of the bones in the
skull, face, arms, and legs. Shown below is a pedigree for a family with Roberts Syndrome.
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A.
Insufficient data
B.
Homozygous recessive
C.
Homozygous dominant
D.
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RNA analysis of wheat genes. Engineers from the Department of Crop and Soil Sciences at
Washington State University used regression to estimate the number of copies of a gene
transcript in an aliquot of ribonucleic acid (RNA) extracted from a wheat plant (Electronic
Journal of Biotechnology, April 15, 2004). The proportion (X1) of RNA extracted from a wheat
plant exposed to the cold was varied, and the transcript copy number (Y , in thousands) was
measured for each of two cloned genes: Mn superoxide dismutase (MnSOD) and phospholipase
D (PLD).
(a) Write a hypothesized model for number of copies (Y ) as a function of proportion (X1) of
RNA extracted and gene type (MnSOD or PLD). Assume that proportion of RNA and gene type
interact to affect Y .
(b) Fit the model you wrote in part (a) to the data.
(c) Conduct a test to determine whether, in fact, proportion of RNA and gene type
interact. Test using = 0.01.
(d) Use the results from part (b) to estimate the rate of increase of number of copies (Y ) with
proportion of RNA extracted for the MnSOD gene type.
Use data below:
RNA
Copies
Gene
1
0
401
MSOD
2
0
336
MSOD
3
0
337
MSOD
4
0.33
711
MSOD
5
0.33
637
MSOD
6
0.33
602
MSOD
7
0.5
985
MSOD
8
0.5
650
MSOD
9
0.5
747
MSOD
10
0.67
904
MSOD
11
0.67
1007
MSOD
12
0.67
1047
MSOD
13
0.8
1151
MSOD
14
0.8
1098
MSOD
15
0.8
1061
MSOD
16
1
1261
MSOD
17
1
1272
MSOD
18
1
1256
MSOD
19
0
80
PLD
20
0
83
PLD
21
0
75
PLD
22
0.33
132
PLD
23
0.33
148
PLD
24
0.33
115
PLD
25
0.5
147
PLD
26
0.5
142
PLD
27
0.5
146
PLD
28
0.67
146
PLD
29
0.67
150
PLD
30
0.67
184
PLD
31
0.8
173
PLD
32
0.8
201
PLD
33
0.8
181
PLD
34
1
193
PLD
35
1
187
PLD
36
1
199
PLD
RNA C opies G ene
1 0 401 MSOD
2 0 336 MSOD
3 0 337 MSOD
4 0.33 711 MSOD
5 0.33 637 MSOD
6 0.33 602 MSOD
7 0.5 985 MSOD
8 0.5 650 MSOD
9 0.5 747 MSOD
10 0 .67 904 MSOD
11 0 .67 1007 MSOD
12 0 .67 1047 MSOD
13 0 .8 1151 MSOD
14 0 .8 1098 MSOD
15 0 .8 1061 MSOD
16 1 1261 MSOD
17 1 1272 MSOD
18 1 1256 MSOD
19 0 80 PLD
20 0 83 PLD
21 0 75 PLD
22 0 .33 132 PLD
23 0 .33 148 PLD
24 0 .33 115 PLD
25 0 .5 147 PLD
26 0 .5 142 PLD
27 0 .5 146 PLD
28 0 .67 146 PLD
29 0 .67 150 PLD
30 0 .67 184 PLD
31 0 .8 173 PLD
32 0 .8 201 PLD
33 0 .8 181 PLD
34 1 193 PLD
35 1 187 PLD
36 1 199 PLD.
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Response Steve evaluated some of the most recent testimonials recorded by the company: The
views from the Gold Leaf dome ear we.
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RNA
Copies
Gene
1
0
401
MSOD
2
0
336
MSOD
3
0
337
MSOD
4
0.33
711
MSOD
5
0.33
637
MSOD
6
0.33
602
MSOD
7
0.5
985
MSOD
8
0.5
650
MSOD
9
0.5
747
MSOD
10
0.67
904
MSOD
11
0.67
1007
MSOD
12
0.67
1047
MSOD
13
0.8
1151
MSOD
14
0.8
1098
MSOD
15
0.8
1061
MSOD
16
1
1261
MSOD
17
1
1272
MSOD
18
1
1256
MSOD
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0
80
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0
83
PLD
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0
75
PLD
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PLD
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0.33
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PLD
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115
PLD
25
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147
PLD
26
0.5
142
PLD
27
0.5
146
PLD
28
0.67
146
PLD
29
0.67
150
PLD
30
0.67
184
PLD
31
0.8
173
PLD
32
0.8
201
PLD
33
0.8
181
PLD
34
1
193
PLD
35
1
187
PLD
36
1
199
PLD
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2 0 336 MSOD
3 0 337 MSOD
4 0.33 711 MSOD
5 0.33 637 MSOD
6 0.33 602 MSOD
7 0.5 985 MSOD
8 0.5 650 MSOD
9 0.5 747 MSOD
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11 0 .67 1007 MSOD
12 0 .67 1047 MSOD
13 0 .8 1151 MSOD
14 0 .8 1098 MSOD
15 0 .8 1061 MSOD
16 1 1261 MSOD
17 1 1272 MSOD
18 1 1256 MSOD
19 0 80 PLD
20 0 83 PLD
21 0 75 PLD
22 0 .33 132 PLD
23 0 .33 148 PLD
24 0 .33 115 PLD
25 0 .5 147 PLD
26 0 .5 142 PLD
27 0 .5 146 PLD
28 0 .67 146 PLD
29 0 .67 150 PLD
30 0 .67 184 PLD
31 0 .8 173 PLD
32 0 .8 201 PLD
33 0 .8 181 PLD
34 1 193 PLD
35 1 187 PLD
36 1 199 PLD.
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