Genetics Homework Problem 1 You are mapping genes in a species of f.pdf
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Genetics Homework Problem 1: You are mapping genes in a species of fruit fly. You cross a wild type fly to a fly from a true breeding mutant strain with green eyes (g), short wings (s w), and hairy legs (hl). The F1 progeny all appear wild type (red eyes, long wings, smooth legs). You then testcross a female F1 fly to a male from the mutant strain, and obtain 200 progeny a. What are the genotypes of the two tiles in the test cross?_____ times ______ b. How many progeny of each phenotype would be expected if the genes were assorting independently? c. Can you reject independent assortment at P=0.05? Use statistical evidence to back up your claim. d. Assuming these genes are all linked on the same chromosome, draw a genetic map showing the order and map distances between the three genes. Solution Answer: a). The genotypes of the two flies in the test cross: Gg Swsw Hlhl X gg swsw hlhl b). If the genes were assorting independently, 8 types of phenotypes are expected in equl proportions 1:1:1:1:1:1:1:1 Green, short, hair: 1 Red, long, smooth: 1 Green, long, smooth: 1 Green, short, smooth: 1 Red, short, hairy: 1 Red, long, hairy: 1 Green, long, hairy: 1 Red, short, hairy: 1 c). Phenotype Observed(O) Expected (E) O-E (O-E)2 (O-E)2/E Green, short, hair: 44 25 19 361.00 14.44 Red, long, smooth 42 25 17 289.00 11.56 Green, long, smooth 20 25 -5 25.00 1.00 Green, short, smooth 5 25 -20 400.00 16.00 Red, short, hairy 20 25 -5 25.00 1.00 Red, long, hairy 7 25 -18 324.00 12.96 Green, long, hairy 30 25 5 25.00 1.00 Red, short, hairy 32 25 7 49.00 1.96 200 200 59.92 Chi-square value = 59.92 The independent assortment at P=0.05 is rejected. d). Explanation Shorthand Number of progeny g sw hl 44 G Sw Hl 42 g Sw Hl 20 g sw Hl 5 G sw hl 20 G Sw hl 7 g Sw hl 30 G sw Hl 32 Imagine the order is G SW Hl / g sw hl 1. If single cross over (SCO) occurs between G & Sw Normal order= G---------Sw & g-----sw After cross over= G-----sw & g------Sw G-----sw recombinants are 20+ 32 = 52 g----- Sw recombinants are 20+ 30 = 50 Total recombinants = 102 RF = (102/200.)*100 =51% 2. If single cross over (SCO) occurs between Sw & Hl Normal order= Sw---------Hl & sw----hl After cross over= Sw-----hl & sw------Hl Sw-----hl recombinants are 30 + 7 = 37 sw------Hl recombinants are 32+5 = 37 Total recombinants = 74 RF = (74/200)*100 = 37% 3. If single cross over (SCO) occurs between G & Hl Normal order= G---------Hl & g------hl After cross over= G-----hl & g------Hl G-----hl recombinants are 20 + 7=27 g------Hl recombinants are 20+5=25 Total recombinants = 52 RF = (52/200)*100 = 26% % RF = Map unit distance The order of genes is ----- G------26 m.u.--------Hl------37 m.u.---------Sw Gene for color of eyes and length of wings are not linked. The gene for legs is linked with gene for wings & gene for eye. Phenotype Observed(O) Expected (E) O-E (O-E)2 (O-E)2/E Green, short, hair: 44 25 19 361.00 14.44 Red, long, smooth 42 25 17 289.00 11.56 Green, long, smooth 20 25 -5 25.00 1.00 Green, sho.
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Wythoff construction and l1-embedding
In a polytope P the faces F (from vertex to facets) define the combinatorics of P. In particular a flag F0, F1, ...., F(n-1) with Fi being a face of dimension i and Fi a subset of F(i+1).
From such a flag system and a subset S of {1,....,n} we can define a new flag system named the Wythoff construction. We consider the l1-embedding of the obtained graphs. We also expose an application of the Wythoff construction to the computation of homology of the Mathieu group M24.
GenotypeObservedType of GameteY F CV.docx
Genotype
Observed
Type of Gamete
Y F CV
310
Parental
y f cv
260
Parental
Y f cv
70
Single-crossover I
Y F CV
80
Single-crossover I
Y F cv
20
Double-crossover
y f CV
10
Double-crossover
Y f CV
135
Single-crossover II
y F cv
115
Single-crossover II
Total
1000
Following is the published distances for three genes
1st
middle
3rd
13.7 mu
43 mu
Make up Data_Sec 06
Recombination and Linkage
A Three point test cross in Drosophila
Linkage
Relationship between two or more genes near each other on the same chromosome is known as Linkage
Law of independent assortment applies to genes on different chromosomes(non linked genes)
Linkage
In a three point cross if there is no linkage what would be the expected ratio for all classes ?
1:1:1:1:1:1:1:1
Aim for three point linkage experiment
Locate the respective position of the three linked gene loci on the same chromosome
Measure distances between genes
Construct the genetic map which estimate the physical distances between gene loci
Drosophila Genes & Phenotypes
Genotype (alleles)
y cv f mutant and + + + (Y CV F) wildtype
Phenotype (appearance)
Mutant: wild type:
y: yellow body grey body
cv: crossveinless wings crossvein wings
f: forked bristles straight bristles
Genes are located on the X chromosome
Parental Cross:
+++ / Y X ycvf / ycvf
Males Females
Phenotype (P): [WT] [yellow, crossveinless, forked]
F1 Generation:
ycvf / Y and +++ / ycvf
Males Females
P: [yellow, crossveinless, forked] [WT]
F1 trihybrid test cross:
ycvf / Y X +++ / ycvf
4 Males 6 Females
First day lab set up
Test Cross-Testing genotype of parents showing dominant phenotype
Test cross- Cross between F1 progeny and recessive parents.
F1 trihybrid test cross:
ycvf / Y x +++ / ycvf
4 Males 6 Females
During meiosis of spermatogenesis male Drosophila flies do not exhibit crossing over between homologous pair - no recombinant product.
X chromosome and Y chromosome do have regions of pairing, and do form tetrads, but don’t recombine.
Male show complete linkage, advantageous for our experiment.
Tri-hybrid females had crossing over .
8 Phenotypes You will be expecting today!!
Grey, crossvein,straight (+ + + )
Yellow,crossveinless,forked(y cv f )
Grey,crossveinless,forked (+cv f )
Yellow,crossvein,straight(y + +)
Grey,crossvein,forked (+ + f)
Yellow,crossveinless,straight (y cv +)
Grey, crossveinless,straight(+cv +)
Yellow,crossvein,forked(y + f)
Place 10 -12 flies at a time to get the phenotypes.
Non-recombinant
Single CO- region I
Single CO-region II
Double CO
Our results
pool out the whole class data to calculate the genetic distanc ...
Statistics Assignment 1 HET551 – Design and Developm.docx
Statistics Assignment 1
HET551 – Design and Development Project 1
Michael Allwright
Haddon O’Neill
Tuesday, 24 May 2011
1 Normal Approximation to the Binomial Distribution
This section of the assignment shows how a normal curve can be used to approximate the binomial distribution. This
section of the assignment was completed using a MATLAB function (shown in Listings 1) which would generate and
save plots of the various Binomial Distributions after normalisation, and then calculate the errors between the standard
normal curve and the binomial distribution.
The plots in Figures 1 and 2 show the binomial distribution for various n trials with probability p = 1
3
and p = 1
2
respectively. These binomial plots have been normalised so that they can be compared with the standard normal
distribution.
From these plots it can be seen that once the binomial distribution has been normalised, the normal approximation is
a good approach to estimating the binomial distribution. To determine its accuracy, the data in Table 1 shows the
evaluation of qn = P(bn ≥ µn + 2σn) for both the normal curve and binomial distribution.
qn = P(bn ≥ µn + 2σn) Calculation Error
n N(0, 1) B(n, 1
2
) B(n, 1
3
) B(n, 1
2
) B(n, 1
3
)
1 0.0228 0.0000 0.0000 -0.02278 -0.02278
2 0.0228 0.0000 0.0000 -0.02278 -0.02278
3 0.0228 0.0000 0.0370 -0.02278 0.01426
4 0.0228 0.0000 0.0123 -0.02278 -0.01043
5 0.0228 0.0313 0.0453 0.00847 0.02249
10 0.0228 0.0107 0.0197 -0.01203 -0.00312
20 0.0228 0.0207 0.0376 -0.00208 0.01486
30 0.0228 0.0214 0.0188 -0.00139 -0.00398
40 0.0228 0.0192 0.0214 -0.00354 -0.00134
50 0.0228 0.0164 0.0222 -0.00636 -0.00059
100 0.0228 0.0176 0.0276 -0.00518 0.00479
Table 1: Calculating the error of the normal approximation to the binomial for various n and p
2 Analytical investigation of the Exponential Distribution
For this part of the assignment the density function shown in Equation 1 was given.
f(x) = λe−λx for x ≥ 0 and λ ≥ 0 (1)
Before any calculations were attempted, the area under graph was checked to show that
´∞
−∞f(x) dx = 1. That is
that the total probability of all possible values was 1.
2.1 Derivation of CDF
To find the CDF of the given function, the function was integrated with 0 and x being the lower and upper bound
respectively. This derivation is shown in Equations 2 to 4.
CDF =
ˆ x
o
f(x) dx =
ˆ x
o
λe−λx dx (2)
2
−5 −4 −3 −2 −1 0 1 2 3 4 5
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
Number of Successes (shifted left by u = 0.33)
P
ro
b
a
b
ili
ty
(a) n = 1
−5 −4 −3 −2 −1 0 1 2 3 4 5
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
Number of Successes (shifted left by u = 0.67)
P
ro
b
a
b
ili
ty
(b) n = 2
−5 −4 −3 −2 −1 0 1 2 3 4 5
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
Number of Successes (shifted left by u = 1.00)
P
ro
b
a
b
ili
ty
(c) n = 3
−5 −4 −3 −2 −1 0 1 2 3 4 5
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
Number of Successes.
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.
SAY. Write your answer in the space provided or on a separate sheet o.pdf
SAY. Write your answer in the space provided or on a separate sheet of pap In the fruit fly.
Drarephila melanu raster, a spineless (no wing a male that is claret bristles) female ny mated to
(dark eyes) and hairless (no th otypically wild-type F1 f nai progeny were mated to fully
homozygous (mutant) males, and the following progeny (1000 total) were observed, claret,
spineless 18 309 claret, hairless hairless, claret, spineless 140 hairless hairless spineless (a)
Which gene is in the middle? (b) With respect to the three genes mentioned in the problem, what
are the genotypes of the homozygous Parents used in making the wild F1 heteroaygote? (c What
are the map distances between the three genes? (d) What is the coefficient of coincidence?
Solution
Answer:
(1) (a) Spineless = (s + +) [Parental]
Claret hairless = (+ c h) [Parental]
Claret = =(+ c +) [double cross over]
hairless,spineless = (s + h) [double cross over]
Therefore, Parentals = s + + , + c h
and Double crossovers = + c + , s + h
Analysing, we get: s+ s+, +c +c and h+ +h.
h is different, so, h (hairless) is in the middle.
(b) The genotypes of the homozygous parents are = s + + / s + + and + c h / + c h
(c) The map distances have been calculated as under:
(Spineless – hairless) = (Number of recombinants)/1000) * 100
= (32 + 38 + 12 + 18)/(1000) * 100
= (100/1000) * 100 = 10% = 10 map units (mu)
(hairless-claret) = (140 + 130 + 12 + 18)/1000) * 100
= (300/1000) * 100
= 30%
= 30 map units (mu)
(d) Coefficient of coincidence = observed double crossovers/expected double crossovers
So, we need to calculate expected double cross overs:
Expected double cross overs = recombination frequency in region 1 (map units / 100) X
recombination frequency in region 2 (map units / 100)
= .(10/100) * (30/100)
= 0.1 * 0.3
= 0.03
Therefore, Coefficient of coincidence = [(18 + 12)/1000] / 0.03
= (30/1000) / 0.03
= 0.03 / 0.03 = 1.
New Families of Odd Harmonious Graphs
In this paper, we show that the number of edges for any odd harmonious Eulerian graph is congruent to 0 or 2 (mod 4), and we found a counter example for the inverse of this statement is not true. We also proved that, the graphs which are constructed by two copies of even cycle Cn sharing a common edge are odd harmonious. In addition, we obtained an odd harmonious labeling for the graphs which are constructed by two copies of cycle Cn sharing a common vertex when n is congruent to 0 (mod 4). Moreover, we show that, the Cartesian product of cycle graph Cm and path Pn for each n ≥ 2, m ≡ 0 (mod 4) are odd harmonious graphs. Finally many new families of odd harmonious graphs are introduced.
Recommended
Wythoff construction and l1-embedding
In a polytope P the faces F (from vertex to facets) define the combinatorics of P. In particular a flag F0, F1, ...., F(n-1) with Fi being a face of dimension i and Fi a subset of F(i+1).
From such a flag system and a subset S of {1,....,n} we can define a new flag system named the Wythoff construction. We consider the l1-embedding of the obtained graphs. We also expose an application of the Wythoff construction to the computation of homology of the Mathieu group M24.
GenotypeObservedType of GameteY F CV.docx
Genotype
Observed
Type of Gamete
Y F CV
310
Parental
y f cv
260
Parental
Y f cv
70
Single-crossover I
Y F CV
80
Single-crossover I
Y F cv
20
Double-crossover
y f CV
10
Double-crossover
Y f CV
135
Single-crossover II
y F cv
115
Single-crossover II
Total
1000
Following is the published distances for three genes
1st
middle
3rd
13.7 mu
43 mu
Make up Data_Sec 06
Recombination and Linkage
A Three point test cross in Drosophila
Linkage
Relationship between two or more genes near each other on the same chromosome is known as Linkage
Law of independent assortment applies to genes on different chromosomes(non linked genes)
Linkage
In a three point cross if there is no linkage what would be the expected ratio for all classes ?
1:1:1:1:1:1:1:1
Aim for three point linkage experiment
Locate the respective position of the three linked gene loci on the same chromosome
Measure distances between genes
Construct the genetic map which estimate the physical distances between gene loci
Drosophila Genes & Phenotypes
Genotype (alleles)
y cv f mutant and + + + (Y CV F) wildtype
Phenotype (appearance)
Mutant: wild type:
y: yellow body grey body
cv: crossveinless wings crossvein wings
f: forked bristles straight bristles
Genes are located on the X chromosome
Parental Cross:
+++ / Y X ycvf / ycvf
Males Females
Phenotype (P): [WT] [yellow, crossveinless, forked]
F1 Generation:
ycvf / Y and +++ / ycvf
Males Females
P: [yellow, crossveinless, forked] [WT]
F1 trihybrid test cross:
ycvf / Y X +++ / ycvf
4 Males 6 Females
First day lab set up
Test Cross-Testing genotype of parents showing dominant phenotype
Test cross- Cross between F1 progeny and recessive parents.
F1 trihybrid test cross:
ycvf / Y x +++ / ycvf
4 Males 6 Females
During meiosis of spermatogenesis male Drosophila flies do not exhibit crossing over between homologous pair - no recombinant product.
X chromosome and Y chromosome do have regions of pairing, and do form tetrads, but don’t recombine.
Male show complete linkage, advantageous for our experiment.
Tri-hybrid females had crossing over .
8 Phenotypes You will be expecting today!!
Grey, crossvein,straight (+ + + )
Yellow,crossveinless,forked(y cv f )
Grey,crossveinless,forked (+cv f )
Yellow,crossvein,straight(y + +)
Grey,crossvein,forked (+ + f)
Yellow,crossveinless,straight (y cv +)
Grey, crossveinless,straight(+cv +)
Yellow,crossvein,forked(y + f)
Place 10 -12 flies at a time to get the phenotypes.
Non-recombinant
Single CO- region I
Single CO-region II
Double CO
Our results
pool out the whole class data to calculate the genetic distanc ...
Statistics Assignment 1 HET551 – Design and Developm.docx
Statistics Assignment 1
HET551 – Design and Development Project 1
Michael Allwright
Haddon O’Neill
Tuesday, 24 May 2011
1 Normal Approximation to the Binomial Distribution
This section of the assignment shows how a normal curve can be used to approximate the binomial distribution. This
section of the assignment was completed using a MATLAB function (shown in Listings 1) which would generate and
save plots of the various Binomial Distributions after normalisation, and then calculate the errors between the standard
normal curve and the binomial distribution.
The plots in Figures 1 and 2 show the binomial distribution for various n trials with probability p = 1
3
and p = 1
2
respectively. These binomial plots have been normalised so that they can be compared with the standard normal
distribution.
From these plots it can be seen that once the binomial distribution has been normalised, the normal approximation is
a good approach to estimating the binomial distribution. To determine its accuracy, the data in Table 1 shows the
evaluation of qn = P(bn ≥ µn + 2σn) for both the normal curve and binomial distribution.
qn = P(bn ≥ µn + 2σn) Calculation Error
n N(0, 1) B(n, 1
2
) B(n, 1
3
) B(n, 1
2
) B(n, 1
3
)
1 0.0228 0.0000 0.0000 -0.02278 -0.02278
2 0.0228 0.0000 0.0000 -0.02278 -0.02278
3 0.0228 0.0000 0.0370 -0.02278 0.01426
4 0.0228 0.0000 0.0123 -0.02278 -0.01043
5 0.0228 0.0313 0.0453 0.00847 0.02249
10 0.0228 0.0107 0.0197 -0.01203 -0.00312
20 0.0228 0.0207 0.0376 -0.00208 0.01486
30 0.0228 0.0214 0.0188 -0.00139 -0.00398
40 0.0228 0.0192 0.0214 -0.00354 -0.00134
50 0.0228 0.0164 0.0222 -0.00636 -0.00059
100 0.0228 0.0176 0.0276 -0.00518 0.00479
Table 1: Calculating the error of the normal approximation to the binomial for various n and p
2 Analytical investigation of the Exponential Distribution
For this part of the assignment the density function shown in Equation 1 was given.
f(x) = λe−λx for x ≥ 0 and λ ≥ 0 (1)
Before any calculations were attempted, the area under graph was checked to show that
´∞
−∞f(x) dx = 1. That is
that the total probability of all possible values was 1.
2.1 Derivation of CDF
To find the CDF of the given function, the function was integrated with 0 and x being the lower and upper bound
respectively. This derivation is shown in Equations 2 to 4.
CDF =
ˆ x
o
f(x) dx =
ˆ x
o
λe−λx dx (2)
2
−5 −4 −3 −2 −1 0 1 2 3 4 5
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
Number of Successes (shifted left by u = 0.33)
P
ro
b
a
b
ili
ty
(a) n = 1
−5 −4 −3 −2 −1 0 1 2 3 4 5
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
Number of Successes (shifted left by u = 0.67)
P
ro
b
a
b
ili
ty
(b) n = 2
−5 −4 −3 −2 −1 0 1 2 3 4 5
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
Number of Successes (shifted left by u = 1.00)
P
ro
b
a
b
ili
ty
(c) n = 3
−5 −4 −3 −2 −1 0 1 2 3 4 5
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
Number of Successes.
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.
SAY. Write your answer in the space provided or on a separate sheet o.pdf
SAY. Write your answer in the space provided or on a separate sheet of pap In the fruit fly.
Drarephila melanu raster, a spineless (no wing a male that is claret bristles) female ny mated to
(dark eyes) and hairless (no th otypically wild-type F1 f nai progeny were mated to fully
homozygous (mutant) males, and the following progeny (1000 total) were observed, claret,
spineless 18 309 claret, hairless hairless, claret, spineless 140 hairless hairless spineless (a)
Which gene is in the middle? (b) With respect to the three genes mentioned in the problem, what
are the genotypes of the homozygous Parents used in making the wild F1 heteroaygote? (c What
are the map distances between the three genes? (d) What is the coefficient of coincidence?
Solution
Answer:
(1) (a) Spineless = (s + +) [Parental]
Claret hairless = (+ c h) [Parental]
Claret = =(+ c +) [double cross over]
hairless,spineless = (s + h) [double cross over]
Therefore, Parentals = s + + , + c h
and Double crossovers = + c + , s + h
Analysing, we get: s+ s+, +c +c and h+ +h.
h is different, so, h (hairless) is in the middle.
(b) The genotypes of the homozygous parents are = s + + / s + + and + c h / + c h
(c) The map distances have been calculated as under:
(Spineless – hairless) = (Number of recombinants)/1000) * 100
= (32 + 38 + 12 + 18)/(1000) * 100
= (100/1000) * 100 = 10% = 10 map units (mu)
(hairless-claret) = (140 + 130 + 12 + 18)/1000) * 100
= (300/1000) * 100
= 30%
= 30 map units (mu)
(d) Coefficient of coincidence = observed double crossovers/expected double crossovers
So, we need to calculate expected double cross overs:
Expected double cross overs = recombination frequency in region 1 (map units / 100) X
recombination frequency in region 2 (map units / 100)
= .(10/100) * (30/100)
= 0.1 * 0.3
= 0.03
Therefore, Coefficient of coincidence = [(18 + 12)/1000] / 0.03
= (30/1000) / 0.03
= 0.03 / 0.03 = 1.
New Families of Odd Harmonious Graphs
In this paper, we show that the number of edges for any odd harmonious Eulerian graph is congruent to 0 or 2 (mod 4), and we found a counter example for the inverse of this statement is not true. We also proved that, the graphs which are constructed by two copies of even cycle Cn sharing a common edge are odd harmonious. In addition, we obtained an odd harmonious labeling for the graphs which are constructed by two copies of cycle Cn sharing a common vertex when n is congruent to 0 (mod 4). Moreover, we show that, the Cartesian product of cycle graph Cm and path Pn for each n ≥ 2, m ≡ 0 (mod 4) are odd harmonious graphs. Finally many new families of odd harmonious graphs are introduced.
4th Semester Electronic and Communication Engineering (2013-June) Question Pa...
4th Semester Electronic and Communication Engineering (2013-June) Question Pa...BGS Institute of Technology, Adichunchanagiri University (ACU)
(Theory)1 A statistical hypothesis is (TFTFTF.docx
(
Theory
)
1 A statistical hypothesis is
(
T
F
T
F
T
F
)a random variable a pivotal quantity
an assertion about unknown distribution of one or more random variables
2 (
∼
)Given arandom variable(absolutely)continue X F(x),F(x)unknown,siaX1,...,Xna sample from XandFn(x)the empirical distribution function.The Kolmogorov-Smirnovtest is usedtoverify
(
T
F
T
F
T
F
)the hypothesis H0 : X ∼ F0 vs Ha : H0 is false, where F0(x) is the hypothesized distribution for X the hypothesis H0 : X hasthe Kolmogorov distribution vs Ha : X hasthe Smirnov distribution the hypothesis H0 : X ∼ F0 vs Ha : H0 is false, using the test statistic sup |Fn(x) − F0(x)|
3 (
x
∈
R
) (
TV
F
TV
F
TV
F
TV
F
)Ina hypothesis testthep−valueisthe significance 1−γ
the value of test statistic
the probabilityof rejecting the true hypothesis
the smallest significancewith which you couldreject H0 when it’s true
4 (
TV
F
TV
F
TV
F
)In a hypothesis testthesignificance level αisthe value of test statistic
the probabilitywith which you could reject H0 when H0 is true
the probabilitywith which you could accept H0 when H0 is false
5
(
),
)For a given experimentare only and only possible,sincompatible resultsSi(i=1,...,sand s≥2), each one of hypothesized probabilityπi=P[Si]; .Repeating n times the experiment the result Si will be observed Ni times. Obviously . Let ni the observed value of Ni and let the chi-squared with s-1 degrees of freedom. Answer correctly.
(
TV
F
)the values represent the observed frequencies of the result Si
(
TV
F
)the statistic is approximately normal when n is big
TV
F
(
TV
F
)the condition is sufficient to ensure that
(
pag.
8
)
6 Given two hypothesis test simple,T1andT2to decide between the null hypothesis Hoand the alternativeHa,with error probability of 1stand 2ndtype respectively equal to α1, β1 for T1 and α2, β2 forT2, then T1 will be preferable to T2 if:
(
TV
F
TV
F
TV
F
TV
F
)α1 <α2 e β1 <β2 α1 = α2 e β1 >β2 α1 = α2 e β1 <β2 α1 >α2 e β1 <β2
(
Part B
)
1. Let U1 and U2 twouniform random variables on [0; 1] and independent.
1.1 To find the joint densityfU(u1,u2)of the vectorU=(U1,U2)Tand to draw the graph.
1.2 LetT=U1+U2.We denote by FT(t)the cumulative function of T.To show that, if 0≤t≤1,then FT(t)is the volume of parallelepiped in blue drawn in the picture. Calculate this volume.
Hint:Calculate P[U1+U2≤t]whereas the points(u1,u2)favorable are such that
{(u1,u2):0≤u1≤1;0≤u2≤1:0≤ u1 + u2 ≤t}.
1.3 Now, it will be easy to show that:
Calculate the probability that
Let now
0.523 0.556 0.994 1.000 1.052 1.087 1.154 1.322 1.485 1.549 (1)
10 observations from an unknown population X.
1.4 To write the sample distribution function Fn(x)concerning the observations(1).
We want to decide about the test (non parametric)
H0:FX(x)=FT(x) Ha : FX(x)
Using the techniques of Kolmogorov-Smirnov.
≠ FT(x) (2)
1.5 At this scope,calledxk; withk=1,...,10the ...
C2.0 propositional logic
I use the slide from internet but I cant get the source just for use .....it is not my work
Recombination and LinkageA Three point test cross in Drosophil.docx
Recombination and Linkage
A Three point test cross in Drosophila
Linkage
Relationship between two or more genes near each other on the same chromosome is known as Linkage
Law of independent assortment applies to genes on different chromosomes(non linked genes)
Linkage
In a three point cross if there is no linkage what would be the expected ratio for all classes ?
1:1:1:1:1:1:1:1
Aim for three point linkage experiment
Locate the respective position of the three linked gene loci on the same chromosome
Measure distances between genes
Construct the genetic map which estimate the physical distances between gene loci
Drosophila Genes & Phenotypes
Genotype (alleles)
y cv f mutant and + + + (Y CV F) wildtype
Phenotype (appearance)
Mutant: wild type:
y: yellow body grey body
cv: crossveinless wings crossvein wings
f: forked bristles straight bristles
Genes are located on the X chromosome
Parental Cross:
+++ / Y X ycvf / ycvf
Males Females
Phenotype (P): [WT] [yellow, crossveinless, forked]
F1 Generation:
ycvf / Y and +++ / ycvf
Males Females
P: [yellow, crossveinless, forked] [WT]
F1 trihybrid test cross:
ycvf / Y X +++ / ycvf
4 Males 6 Females
First day lab set up
Test Cross-Testing genotype of parents showing dominant phenotype
Test cross- Cross between F1 progeny and recessive parents.
F1 trihybrid test cross:
ycvf / Y x +++ / ycvf
4 Males 6 Females
During meiosis of spermatogenesis male Drosophila flies do not exhibit crossing over between homologous pair - no recombinant product.
X chromosome and Y chromosome do have regions of pairing, and do form tetrads, but don’t recombine.
Male show complete linkage, advantageous for our experiment.
Tri-hybrid females had crossing over .
8 Phenotypes You will be expecting today!!
Grey, crossvein,straight (+ + + )
Yellow,crossveinless,forked(y cv f )
Grey,crossveinless,forked (+cv f )
Yellow,crossvein,straight(y + +)
Grey,crossvein,forked (+ + f)
Yellow,crossveinless,straight (y cv +)
Grey, crossveinless,straight(+cv +)
Yellow,crossvein,forked(y + f)
Place 10 -12 flies at a time to get the phenotypes.
Non-recombinant
Single CO- region I
Single CO-region II
Double CO
Our results
pool out the whole class data to calculate the genetic distance more accurately
Today’s Goal: Construct a crude genetic map
Find the gene position
calculate c.o.c., interference
Perform X²(to determine if the map distance from the pooled class data are statistically same with the published map distance)
F2 Trihybrid Test Cross example
axp / +++ X axp / Y
gametes a x p ...
E6 GUT Model and the Higgs boson search
E6 GUT model & the
Higgs boson search
Gökhan Ünel / UC Irvine & CERN
Saleh Sultansoy / Ankara
11 October 2006
Flavour at the LHC workshop
JAVALAB #8 - ARRAY BASED LISTSThe next exercise is based on this.pdf
JAVA
LAB #8 - ARRAY BASED LISTS
The next exercise is based on this implemetation for an UnorderedArrayList of integers:
//Interface: ArrayListADT
//works for int
public interface ArrayListADT {
public boolean isEmpty(); //Method to determine whether the list is empty.
public boolean isFull(); //Method to determine whether the list is full.
public int listSize(); //Method to return the number of elements in the list.
public int maxListSize(); //Method to return the maximum size of the list.
public void print(); //Method to output the elements of the list.
public boolean isItemAtEqual(int location, int item); //Method to determine whether item is
the same as the item in the list at location.
public void insertAt(int location, int insertItem); //Method to insert insertItem in the list at
the position
public void insertEnd(int insertItem); //Method to insert insertItem at the end of the list.
public void removeAt(int location); //Method to remove the item from the list at location.
public int retrieveAt(int location); //Method to retrieve the element from the list at location.
public void replaceAt(int location, int repItem); //Method to replace the element in the list at
location with repItem.
public void clearList(); //Method to remove all the elements from the list.
public int search(int searchItem); //Method to determine whether searchItem is in the list.
public void remove(int removeItem); //Method to remove an item from the list.
}
//Class: ArrayListClass implements
//Interface: ArrayListADT
public abstract class ArrayListClass implements ArrayListADT {
protected int length; //to store the length of the list
protected int maxSize; //to store the maximum size of the list
protected int[] list; //array to hold the list elements
//Default constructor
public ArrayListClass() {
maxSize = 100;
length = 0;
list = new int[maxSize];
}
//Alternate Constructor
public ArrayListClass(int size) {
if(size <= 0) {
System.err.println(\"The array size must be positive. Creating an array of size 100.\");
maxSize = 100;
}
else
maxSize = size;
length = 0;
list = new int[maxSize];
}
public boolean isEmpty() {
return (length == 0);
}
public boolean isFull() {
return (length == maxSize);
}
public int listSize() {
return length;
}
public int maxListSize() {
return maxSize;
}
public void print() {
for (int i = 0; i < length; i++)
System.out.print(list[i] + \" \");
System.out.println();
}
public boolean isItemAtEqual(int location, int item) {
if (location < 0 || location >= length) {
System.err.println(\"The location of the item to be compared is out of range.\");
return false;
}
return list[location]== item;
}
public void clearList() {
for (int i = 0; i < length; i++)
list[i] = 0;
length = 0;
System.gc(); //invoke the Java garbage collector
}
public void removeAt(int location) {
if (location < 0 || location >= length)
System.err.println(\"The location of the item to be removed is out of range.\");
else {
for(int i = location; i < length - 1; i++)
list[i] = list[i + 1.
In the year 2000, a halophile was isolated from a 250 million year o.pdf
In the year 2000, a halophile was isolated from a 250 million year old salt crystal and grown up
in a university laboratory in Westchester PA. What did the scientists isolate and why do you
think it survived 250 million years?
Solution
Halophile is salt loving, They are extremophiles which live in high saline environment. They
adapt different strategies to survive in the extreme salt conditions. These adaptations are-
Hence the halophile survived 250 million years.
In in the image seen here, in which direction would you expect water .pdf
In in the image seen here, in which direction would you expect water to move if the membrane
is not permeable to the solute? It would move equally between both sides. Right to left Left to
right
Solution
The process shown here is called as Osmosis.. According to which the movement of water
molecules take place from the region higher concentration to a region of lower water
concentration when separated by the semi - permeable Membrane..
In this the left side contains more water molecules concentration then right side as the right side
contains more solute molecules...
Soo, the movement will be from Left Side to Right side...!!!!.
How does the movement of DNA fragments on the gel and the movemen.pdf
How does the movement of DNA fragments on the gel and the movement of the ethilisim
troemide in the eel affect the appearance of the DNA fragments?
Solution
Answer:
the movement of DNA happens from a negative charged electrode to positively charged
electrode as the DNA is negatively charged.
By applying the electric field through the agarose gel matrix DNA is seperated on the basis of
size.
Ethidium bromide is a flourescent dye which intercalates into circular DNA and can
sunsequently change the charge, length, as well as the superhelicity of the DNA molecule,
EtBr presence in gel during electrophoresis can affect movement of DNA. Agarose gel
electrophoresis can be used to resolve circular DNA with different supercoiling topology.
as more EtBr binds between the DNA bases, the molecule stiffens and hence mobility is
decreased as compared to same frangment being run without EtBr.
Ethedium bromide is positively charged and runs from the positive electrode towards the
negative electrode. i.e opposite direction of movement of DNA.
hence when the two encounter, the intercalaton happens, leading to flourescence of orange red
colour under UV light..
For f(x) =2x, find a formula for the Riemann sum obtained by dividin.pdf
For f(x) =2x, find a formula for the Riemann sum obtained by dividing the interval [2.5]
subintervals and using the right hand endpoint for each ck. Simplify the sum and take the limit as
n--> infinity to calculate the area under the curve over [2,5]
please show all of your work as be as descriptive as you can I appreciate your help thank you!
Solution
a=2, b =5, x =(b-a)/n =(5-2)/n =3/n
f(x)=2x
xi=a+ ix =2+i(3/n)
f(xi)=2(2+3(i/n))
the area under the curve over [2,5] is =limn->[i =1 to n]f(xi)x
=limn->[i =1 to n]2(2+3(i/n))(3/n)
=limn->[i =1 to n]6((2/n)+3(i/n2))
=limn-> 6((2n/n)+3(n(n+1)/2n2))
=limn-> 6(2 +3((1+(1/n))/2))
= 6(2 +3((1+0)/2))
=6(2 +3(1/2))
=6(2 +(3/2))
=6(7/2)
=3*7
=21
the area under the curve over [2,5] is 21.
Fibroblasts are the typical cells of simple squamous epithelium. Dens.pdf
Fibroblasts are the typical cells of simple squamous epithelium. Dense connective tissue, cardiac
muscle tissue. skeletal muscle tissue. nervous tissue.
Solution
Answer :- option B
Explanation:- Fibroblast are the type of cells that synthesise extracellular matrix and collagen
and are part of connective tissue..
Explain the difference between an observational study and a designed.pdf
Explain the difference between an observational study and a designed experiment?
Solution
An observational study simply take down the properties of its subjects, without manipulating
them or interfering with their activities.
In a designed experiment, the researcher may change the environment or circumstances of some
of the sample in order to compare, say, a control and a treatment group..
Discuss the recruitment and selection procedure done by human resour.pdf
Discuss the recruitment and selection procedure done by human resource management?
Solution
The procedure of recruitment and selection done by the human resource management involves
five distinct and basic aspects which are as follows:
1. CRITERIA DEVELOPMENT: Those individuals which are involved in the process of hiring
should be trained enough to develop criteria , to conduct interview , to review the resumes , to
develop the questions for interview , to weigh the candidates.
In this step the basic thing is to plan the interview and develop the criteria as which of the
sources are to be used during interview and how to use these sources to score better in the
interview.The criteria should be job focussed and also the focus should be on the personality.
In this process the involvement of discussion of the skills, abilities and some personal
characteristics takes place that are required to have to the job.Before reviewing the resumes by
developing this criteria can be sure that he is fair in selecting the applicants to interview.The
application should include the details of the candidate , previous job experience certificates and
educational qualification certificates.
2. APPLICATION AND REVIEWING OF RESUME : After the completion of step 1 the
applications can now be reviewed. There are different ways to go through this process but there
are various computer based programs that can search for the keywords in the resumes and
shortlist the number of resumes that can b looked into.
3. INTERVIEW PROCESS : After shortlisting those applicants who have met the minimum
criteria.The HR manager selects these applicants for the interview.Due to lack of time sometimes
interview is being done on phone.
4. TEST ADMINISTRATION : Before the hiring decision is made any number of test are
administered. The tests may be physical tests ,drug tests, response test, credit report checks,
personality tests, etc.
5. OFFER MAKING PROCESS : This is the final step in which the choosen candidate is offered
the job position. The offer can b developed through email or through the job letter.All the
benefits and compensation are discussed in the offer letter..
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(Theory)1 A statistical hypothesis is (TFTFTF.docx
(
Theory
)
1 A statistical hypothesis is
(
T
F
T
F
T
F
)a random variable a pivotal quantity
an assertion about unknown distribution of one or more random variables
2 (
∼
)Given arandom variable(absolutely)continue X F(x),F(x)unknown,siaX1,...,Xna sample from XandFn(x)the empirical distribution function.The Kolmogorov-Smirnovtest is usedtoverify
(
T
F
T
F
T
F
)the hypothesis H0 : X ∼ F0 vs Ha : H0 is false, where F0(x) is the hypothesized distribution for X the hypothesis H0 : X hasthe Kolmogorov distribution vs Ha : X hasthe Smirnov distribution the hypothesis H0 : X ∼ F0 vs Ha : H0 is false, using the test statistic sup |Fn(x) − F0(x)|
3 (
x
∈
R
) (
TV
F
TV
F
TV
F
TV
F
)Ina hypothesis testthep−valueisthe significance 1−γ
the value of test statistic
the probabilityof rejecting the true hypothesis
the smallest significancewith which you couldreject H0 when it’s true
4 (
TV
F
TV
F
TV
F
)In a hypothesis testthesignificance level αisthe value of test statistic
the probabilitywith which you could reject H0 when H0 is true
the probabilitywith which you could accept H0 when H0 is false
5
(
),
)For a given experimentare only and only possible,sincompatible resultsSi(i=1,...,sand s≥2), each one of hypothesized probabilityπi=P[Si]; .Repeating n times the experiment the result Si will be observed Ni times. Obviously . Let ni the observed value of Ni and let the chi-squared with s-1 degrees of freedom. Answer correctly.
(
TV
F
)the values represent the observed frequencies of the result Si
(
TV
F
)the statistic is approximately normal when n is big
TV
F
(
TV
F
)the condition is sufficient to ensure that
(
pag.
8
)
6 Given two hypothesis test simple,T1andT2to decide between the null hypothesis Hoand the alternativeHa,with error probability of 1stand 2ndtype respectively equal to α1, β1 for T1 and α2, β2 forT2, then T1 will be preferable to T2 if:
(
TV
F
TV
F
TV
F
TV
F
)α1 <α2 e β1 <β2 α1 = α2 e β1 >β2 α1 = α2 e β1 <β2 α1 >α2 e β1 <β2
(
Part B
)
1. Let U1 and U2 twouniform random variables on [0; 1] and independent.
1.1 To find the joint densityfU(u1,u2)of the vectorU=(U1,U2)Tand to draw the graph.
1.2 LetT=U1+U2.We denote by FT(t)the cumulative function of T.To show that, if 0≤t≤1,then FT(t)is the volume of parallelepiped in blue drawn in the picture. Calculate this volume.
Hint:Calculate P[U1+U2≤t]whereas the points(u1,u2)favorable are such that
{(u1,u2):0≤u1≤1;0≤u2≤1:0≤ u1 + u2 ≤t}.
1.3 Now, it will be easy to show that:
Calculate the probability that
Let now
0.523 0.556 0.994 1.000 1.052 1.087 1.154 1.322 1.485 1.549 (1)
10 observations from an unknown population X.
1.4 To write the sample distribution function Fn(x)concerning the observations(1).
We want to decide about the test (non parametric)
H0:FX(x)=FT(x) Ha : FX(x)
Using the techniques of Kolmogorov-Smirnov.
≠ FT(x) (2)
1.5 At this scope,calledxk; withk=1,...,10the ...
C2.0 propositional logic
I use the slide from internet but I cant get the source just for use .....it is not my work
Recombination and LinkageA Three point test cross in Drosophil.docx
Recombination and Linkage
A Three point test cross in Drosophila
Linkage
Relationship between two or more genes near each other on the same chromosome is known as Linkage
Law of independent assortment applies to genes on different chromosomes(non linked genes)
Linkage
In a three point cross if there is no linkage what would be the expected ratio for all classes ?
1:1:1:1:1:1:1:1
Aim for three point linkage experiment
Locate the respective position of the three linked gene loci on the same chromosome
Measure distances between genes
Construct the genetic map which estimate the physical distances between gene loci
Drosophila Genes & Phenotypes
Genotype (alleles)
y cv f mutant and + + + (Y CV F) wildtype
Phenotype (appearance)
Mutant: wild type:
y: yellow body grey body
cv: crossveinless wings crossvein wings
f: forked bristles straight bristles
Genes are located on the X chromosome
Parental Cross:
+++ / Y X ycvf / ycvf
Males Females
Phenotype (P): [WT] [yellow, crossveinless, forked]
F1 Generation:
ycvf / Y and +++ / ycvf
Males Females
P: [yellow, crossveinless, forked] [WT]
F1 trihybrid test cross:
ycvf / Y X +++ / ycvf
4 Males 6 Females
First day lab set up
Test Cross-Testing genotype of parents showing dominant phenotype
Test cross- Cross between F1 progeny and recessive parents.
F1 trihybrid test cross:
ycvf / Y x +++ / ycvf
4 Males 6 Females
During meiosis of spermatogenesis male Drosophila flies do not exhibit crossing over between homologous pair - no recombinant product.
X chromosome and Y chromosome do have regions of pairing, and do form tetrads, but don’t recombine.
Male show complete linkage, advantageous for our experiment.
Tri-hybrid females had crossing over .
8 Phenotypes You will be expecting today!!
Grey, crossvein,straight (+ + + )
Yellow,crossveinless,forked(y cv f )
Grey,crossveinless,forked (+cv f )
Yellow,crossvein,straight(y + +)
Grey,crossvein,forked (+ + f)
Yellow,crossveinless,straight (y cv +)
Grey, crossveinless,straight(+cv +)
Yellow,crossvein,forked(y + f)
Place 10 -12 flies at a time to get the phenotypes.
Non-recombinant
Single CO- region I
Single CO-region II
Double CO
Our results
pool out the whole class data to calculate the genetic distance more accurately
Today’s Goal: Construct a crude genetic map
Find the gene position
calculate c.o.c., interference
Perform X²(to determine if the map distance from the pooled class data are statistically same with the published map distance)
F2 Trihybrid Test Cross example
axp / +++ X axp / Y
gametes a x p ...
E6 GUT Model and the Higgs boson search
E6 GUT model & the
Higgs boson search
Gökhan Ünel / UC Irvine & CERN
Saleh Sultansoy / Ankara
11 October 2006
Flavour at the LHC workshop
Similar to Genetics Homework Problem 1 You are mapping genes in a species of f.pdf (14)
4th Semester Electronic and Communication Engineering (2013-June) Question Pa...
4th Semester Electronic and Communication Engineering (2013-June) Question Pa...
(Theory)1 A statistical hypothesis is (TFTFTF.docx
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More from arpaqindia
JAVALAB #8 - ARRAY BASED LISTSThe next exercise is based on this.pdf
JAVA
LAB #8 - ARRAY BASED LISTS
The next exercise is based on this implemetation for an UnorderedArrayList of integers:
//Interface: ArrayListADT
//works for int
public interface ArrayListADT {
public boolean isEmpty(); //Method to determine whether the list is empty.
public boolean isFull(); //Method to determine whether the list is full.
public int listSize(); //Method to return the number of elements in the list.
public int maxListSize(); //Method to return the maximum size of the list.
public void print(); //Method to output the elements of the list.
public boolean isItemAtEqual(int location, int item); //Method to determine whether item is
the same as the item in the list at location.
public void insertAt(int location, int insertItem); //Method to insert insertItem in the list at
the position
public void insertEnd(int insertItem); //Method to insert insertItem at the end of the list.
public void removeAt(int location); //Method to remove the item from the list at location.
public int retrieveAt(int location); //Method to retrieve the element from the list at location.
public void replaceAt(int location, int repItem); //Method to replace the element in the list at
location with repItem.
public void clearList(); //Method to remove all the elements from the list.
public int search(int searchItem); //Method to determine whether searchItem is in the list.
public void remove(int removeItem); //Method to remove an item from the list.
}
//Class: ArrayListClass implements
//Interface: ArrayListADT
public abstract class ArrayListClass implements ArrayListADT {
protected int length; //to store the length of the list
protected int maxSize; //to store the maximum size of the list
protected int[] list; //array to hold the list elements
//Default constructor
public ArrayListClass() {
maxSize = 100;
length = 0;
list = new int[maxSize];
}
//Alternate Constructor
public ArrayListClass(int size) {
if(size <= 0) {
System.err.println(\"The array size must be positive. Creating an array of size 100.\");
maxSize = 100;
}
else
maxSize = size;
length = 0;
list = new int[maxSize];
}
public boolean isEmpty() {
return (length == 0);
}
public boolean isFull() {
return (length == maxSize);
}
public int listSize() {
return length;
}
public int maxListSize() {
return maxSize;
}
public void print() {
for (int i = 0; i < length; i++)
System.out.print(list[i] + \" \");
System.out.println();
}
public boolean isItemAtEqual(int location, int item) {
if (location < 0 || location >= length) {
System.err.println(\"The location of the item to be compared is out of range.\");
return false;
}
return list[location]== item;
}
public void clearList() {
for (int i = 0; i < length; i++)
list[i] = 0;
length = 0;
System.gc(); //invoke the Java garbage collector
}
public void removeAt(int location) {
if (location < 0 || location >= length)
System.err.println(\"The location of the item to be removed is out of range.\");
else {
for(int i = location; i < length - 1; i++)
list[i] = list[i + 1.
In the year 2000, a halophile was isolated from a 250 million year o.pdf
In the year 2000, a halophile was isolated from a 250 million year old salt crystal and grown up
in a university laboratory in Westchester PA. What did the scientists isolate and why do you
think it survived 250 million years?
Solution
Halophile is salt loving, They are extremophiles which live in high saline environment. They
adapt different strategies to survive in the extreme salt conditions. These adaptations are-
Hence the halophile survived 250 million years.
In in the image seen here, in which direction would you expect water .pdf
In in the image seen here, in which direction would you expect water to move if the membrane
is not permeable to the solute? It would move equally between both sides. Right to left Left to
right
Solution
The process shown here is called as Osmosis.. According to which the movement of water
molecules take place from the region higher concentration to a region of lower water
concentration when separated by the semi - permeable Membrane..
In this the left side contains more water molecules concentration then right side as the right side
contains more solute molecules...
Soo, the movement will be from Left Side to Right side...!!!!.
How does the movement of DNA fragments on the gel and the movemen.pdf
How does the movement of DNA fragments on the gel and the movement of the ethilisim
troemide in the eel affect the appearance of the DNA fragments?
Solution
Answer:
the movement of DNA happens from a negative charged electrode to positively charged
electrode as the DNA is negatively charged.
By applying the electric field through the agarose gel matrix DNA is seperated on the basis of
size.
Ethidium bromide is a flourescent dye which intercalates into circular DNA and can
sunsequently change the charge, length, as well as the superhelicity of the DNA molecule,
EtBr presence in gel during electrophoresis can affect movement of DNA. Agarose gel
electrophoresis can be used to resolve circular DNA with different supercoiling topology.
as more EtBr binds between the DNA bases, the molecule stiffens and hence mobility is
decreased as compared to same frangment being run without EtBr.
Ethedium bromide is positively charged and runs from the positive electrode towards the
negative electrode. i.e opposite direction of movement of DNA.
hence when the two encounter, the intercalaton happens, leading to flourescence of orange red
colour under UV light..
For f(x) =2x, find a formula for the Riemann sum obtained by dividin.pdf
For f(x) =2x, find a formula for the Riemann sum obtained by dividing the interval [2.5]
subintervals and using the right hand endpoint for each ck. Simplify the sum and take the limit as
n--> infinity to calculate the area under the curve over [2,5]
please show all of your work as be as descriptive as you can I appreciate your help thank you!
Solution
a=2, b =5, x =(b-a)/n =(5-2)/n =3/n
f(x)=2x
xi=a+ ix =2+i(3/n)
f(xi)=2(2+3(i/n))
the area under the curve over [2,5] is =limn->[i =1 to n]f(xi)x
=limn->[i =1 to n]2(2+3(i/n))(3/n)
=limn->[i =1 to n]6((2/n)+3(i/n2))
=limn-> 6((2n/n)+3(n(n+1)/2n2))
=limn-> 6(2 +3((1+(1/n))/2))
= 6(2 +3((1+0)/2))
=6(2 +3(1/2))
=6(2 +(3/2))
=6(7/2)
=3*7
=21
the area under the curve over [2,5] is 21.
Fibroblasts are the typical cells of simple squamous epithelium. Dens.pdf
Fibroblasts are the typical cells of simple squamous epithelium. Dense connective tissue, cardiac
muscle tissue. skeletal muscle tissue. nervous tissue.
Solution
Answer :- option B
Explanation:- Fibroblast are the type of cells that synthesise extracellular matrix and collagen
and are part of connective tissue..
Explain the difference between an observational study and a designed.pdf
Explain the difference between an observational study and a designed experiment?
Solution
An observational study simply take down the properties of its subjects, without manipulating
them or interfering with their activities.
In a designed experiment, the researcher may change the environment or circumstances of some
of the sample in order to compare, say, a control and a treatment group..
Discuss the recruitment and selection procedure done by human resour.pdf
Discuss the recruitment and selection procedure done by human resource management?
Solution
The procedure of recruitment and selection done by the human resource management involves
five distinct and basic aspects which are as follows:
1. CRITERIA DEVELOPMENT: Those individuals which are involved in the process of hiring
should be trained enough to develop criteria , to conduct interview , to review the resumes , to
develop the questions for interview , to weigh the candidates.
In this step the basic thing is to plan the interview and develop the criteria as which of the
sources are to be used during interview and how to use these sources to score better in the
interview.The criteria should be job focussed and also the focus should be on the personality.
In this process the involvement of discussion of the skills, abilities and some personal
characteristics takes place that are required to have to the job.Before reviewing the resumes by
developing this criteria can be sure that he is fair in selecting the applicants to interview.The
application should include the details of the candidate , previous job experience certificates and
educational qualification certificates.
2. APPLICATION AND REVIEWING OF RESUME : After the completion of step 1 the
applications can now be reviewed. There are different ways to go through this process but there
are various computer based programs that can search for the keywords in the resumes and
shortlist the number of resumes that can b looked into.
3. INTERVIEW PROCESS : After shortlisting those applicants who have met the minimum
criteria.The HR manager selects these applicants for the interview.Due to lack of time sometimes
interview is being done on phone.
4. TEST ADMINISTRATION : Before the hiring decision is made any number of test are
administered. The tests may be physical tests ,drug tests, response test, credit report checks,
personality tests, etc.
5. OFFER MAKING PROCESS : This is the final step in which the choosen candidate is offered
the job position. The offer can b developed through email or through the job letter.All the
benefits and compensation are discussed in the offer letter..
Do The statement of stockholders equity explains changes in equity.pdf
Do The statement of stockholder\'s equity explains changes in equity from net income?
Solution
Statement of equity is a summary of changes in share holders equity during the reporting period.
The statemnt includes all changes including issue of new stocks, payment of dividend, issue of
bonus shares, buy back of shares. It also includes changes to retained earnings, reserves and
stock reserve.
Distribution of dividend from net income and transfer of remaining balance to reserves are
shown at the bottom of P&L account.
The remaining balance of profits transferred to reserves are reflected in Statement of changes in
equity as addition to retained earnings.
Hope I have answered the question, for any further clarification please comment..
Describe what is meant by “radical cure” and identify the drug that .pdf
Describe what is meant by “radical cure” and identify the drug that is produces a radical cure for
malaria.
Solution
Radical cure refers to the elimination of malaria parasites completely from the body.It
specifically refers to the elimination of hypnozoites of P.vivax and P. ovale which are the
parasitic stages in the liver and could result in relapse of the infection. A combination of two
drugs i.e chloroquine and primaquine produces a radical cure for malaria..
Describe the nutritional modes of organisms based upon where they obt.pdf
Describe the nutritional modes of organisms based upon where they obtain carbon energy.
Include an example of each.
Solution
Describe the nutritional modes of organisms based upon where they obtain carbon and energy.
Almost all organisms on the earth need energy to do their work, and carbon to use in building
their chemical structures. And these two (energy and carbon) has to be obtained from external
sources, based on the way they obtain energy and carbon, organisms are broadly classified into
four different groups.
Classification
Obtaining energy and carbon
Examples
Autotrophs
Photoautotrophs
They get energy from sunlight and will not depend on other organisms for carbon. Use CO2 as a
carbon source.
All photosynthetic prokaryotes
Plants (eukaryotes)
Chemoautotrophs
These make use of inorganic chemicals as an energy source. For example, H2S, Fe2+ etc and
CO2 will serve as a carbon source.
Archaea living in hydrothermal vents.
Heterotrophs
Photoheterotrophs
Use light as an energy source and will use organic molecules as a carbon source. These type of
organisms are rare.
Heliobacteria, non-sulfur bacteria, green non-sulfur bacteria etc
Chemoheterotrophs
Use organic molecules like sugars as an energy source.
Use organic molecules like proteins, sugars, fats, nucleic acids, etc. as a carbon source. This type
of nutrition is most common form of nutrition on Earth
Most of the prokaryotes, protists, fungi, and animals
Classification
Obtaining energy and carbon
Examples
Autotrophs
Photoautotrophs
They get energy from sunlight and will not depend on other organisms for carbon. Use CO2 as a
carbon source.
All photosynthetic prokaryotes
Plants (eukaryotes)
Chemoautotrophs
These make use of inorganic chemicals as an energy source. For example, H2S, Fe2+ etc and
CO2 will serve as a carbon source.
Archaea living in hydrothermal vents.
Heterotrophs
Photoheterotrophs
Use light as an energy source and will use organic molecules as a carbon source. These type of
organisms are rare.
Heliobacteria, non-sulfur bacteria, green non-sulfur bacteria etc
Chemoheterotrophs
Use organic molecules like sugars as an energy source.
Use organic molecules like proteins, sugars, fats, nucleic acids, etc. as a carbon source. This
type of nutrition is most common form of nutrition on Earth
Most of the prokaryotes, protists, fungi, and animals.
Decasper and Spence (1986) measured babies sucking rates as they lis.pdf
Decasper and Spence (1986) measured babies sucking rates as they liscenced to a story they had
heard in the womb (the cat in the hat) and to a story they had not heard (the king, the mice, and
the cheese) the study concluded that:
a.) amniotic fluid blocks sound waves from reaching the fetus
b) increased heart rate showed that an infant can learn even before it is born
c) different sucking styles and rates showed that an infant can learn even before it is born
d) the cadence of the reading material not previous experience altered sucking style and rate
Solution
different sucking styles and rates showed that an infant can learn even before it is born.
Corhin Eckles and Connor Hayes have recieved walkie talkies for Chri.pdf
Corhin Eckles and Connor Hayes have recieved walkie talkies for Christmas. If they leave from
the same point at the same time, Corhin walking north at 2.5 mph and Connor walking east at 3
mph, how long will they be able to talk to each other if the range of the walkie-talkies is 4 mi?
Solution
Corhin DATA:
distance= x miles; rate=2.5 mph; time=d/r= x/2.5
Connor DATA:
distance= y miles; rate=3 mph; time=y/4
EQUATIONS:
1) x^2+y^2=4^2
2) x/2.5 = y/4
Using 2) you get y=4x/2.5
Substitute into 1) to get x^2+(4x/2.5)^2=16
x^2+(16x^2/6.25)=16
6.25x^2+16x^2=16(6.25)
22.25x^2=100
x^2=100/22.25=4.49...
x=2.12 hrs.
x/2.5= 0.85 hrs. = 50.87 minutes
This is Corhin\'s time, but it is the same for Connor..
Consider three genes in a hypothetical bird. One controls the color o.pdf
Consider three genes in a hypothetical bird. One controls the color of feathers on the top of the
head (red or brown), one controls the color of feathers on the chest (black or brown) and one
controls the number of syllables in the song (3 or 4). You notice that birds frequently have either
red heads and black chests or brown heads and brown chests, but very few have red heads and
brown chests or brown heads and black chests. However, there doesn\'t seem to be any
association between the syllables in the bird song and color. In other words, red heads are
equally likely to sing songs with 3 or 4 syllables, and the same goes for the brown heads. Based
on what you know about how the genome is organized and passed down to the next generation,
explain this observation.
Solution
The observation can be attributed to the mechanism of genetic linkage.
It is a well known fact that genes get passed from one generation to another. Mendel\'s second
law of inheritance states that during this process, the inheritance of one gene is independent of
the inheritance of another gene. Howeve, this law holds true only when the different genes are
sufficiently spaced apart from each other. Inother words, if the genes are close to each other, it is
difficult for them to be inherited in an independent manner. If one of the genes is inherited, then
the other gene (by virtue of its proximity to the first gene) also tends to get inherited. This
phenomenon is called as genetic linkage, and the genes involved are said to be Linked Genes.
The same mechanism is in action in the above case.
The genes for Red Head and Black Crest are linked, as are the genes for Brown Head and Brown
Crest. Therefore, these two genes are inherited in a linked manner, and not in an independent
manner. The two genes therefore behave as One Single Gene.
The gene for song syllable is at a sufficient distance from the linked genes, and follows
Mendel\'s law of independent assortment and is not dependent on the presence or absence of the
former two traits..
Answer the following questions What is the domain of logarithmic fun.pdf
Answer the following questions: What is the domain of logarithmic function What is the range
of Logarithmic function What is the value of e(approximate to 4 decimal places)
Solution
SOLUTION
===========
the domain of any logarithmic function is positive real numbers
1) Domain =(0,inf)
2) Range =(-inf,+inf)
3) e=2.7183.
are organic compounds that bind very tightly to iron ions for transp.pdf
are organic compounds that bind very tightly to iron ions for transport into cells.
Solution
Organisms like bacteria secrerte substances called siderophores which bind to iron for transport
inside the cell. Iron is one of the important micronutrients for the bacterial metabolism.
Ferrichrome,Desferrioxamine B and Desferrioxamine E are some of the important siderophores
secreted by various bacteria..
Book Business Its Legal, Ethical, Global Environment 10th Edition.pdf
Book: Business It\'s Legal, Ethical, Global Environment 10th Edition
Page 204-205
Question and Problems
# 2
#5
Solution
Ethical issues connect intimately with economic issues. Take the economic practice of doing a
cost-benefit analysis. You could spend one hundred dollars for a night on the town, or you could
donate that one hundred dollars to the reelection campaign of your favorite politician. Which
option is better? The night on the town increases pleasure. A politician’s successful campaign
may lead to more liberty in the long term. We regularly make decisions like this, weighing our
options by measuring their likely costs and likely benefits against each other.
This connects economics directly to a major issue in ethics: By what standard do we determine
what counts as a benefit or a cost? A list of competing candidates for the status of ultimate value
standard includes happiness, satisfying the will of God, long-term survival, liberty, duty, and
equality.
Economists implicitly adopt a value framework when beginning a cost-benefit analysis. Different
value commitments can lead to the same item being considered a cost from one perspective and a
benefit from another. For example, those whose standard of value is increasing human happiness
would count a new road to a scenic mountain vista as a benefit, while those whose standard is
maintaining an unchanged natural environment would count it as a cost.
The results of economic analysis also lead directly to ethical issues. For example, one result of
the nineteenth- and twentieth-century debate over capitalism and socialism is a general
consensus that capitalism is effective at producing wealth and socialism is effective at keeping
people poor. Advocates of capitalism use these results to argue that capitalism is good; others
might respond that “socialism is good in theory, but unfortunately it is not practical.” Implicit in
the capitalist position is the view that practical consequences determine goodness. By contrast,
implicit in the position of those who believe socialism to be an impractical moral ideal is the
view that goodness is distinct from practical consequences.
This connects economics to a second major issue in ethics: Is goodness or badness determined by
real-world practical consequences or by some other means, such as revelations from God, faith in
authorities or authoritative institutions, appeals to rational consistency, felt senses of empathy, or
an innate conscience? The point for economic analysis, most of which is a matter of
understanding and predicting the consequences of various actions, is that the relevance of
economic analysis to policymaking depends, in part, on what one believes is the final source of
value standards..
According the United States (U.S.) Centers for Disease Control and P.pdf
According the United States (U.S.) Centers for Disease Control and Prevention, in 2015
influenza and pneumonia combined were the 8th leading causes of death (55,227 deaths in total)
and the only communicable diseases ranked among the top ten killers in the U.S. What were the
10 causes of mortality in the U.S. in 1900 and 2015? Since communicable disease caused a
relatively few deaths in the U.S. in 2015, why should public and public health professionals be
concerned about them anymore and devote resources in their prevention and control?
Solution
Top 10 leading causes of deaths in US: Hearts diseases Cancer Airway diseases which are non
infectious Cardio vascular disorders Accidents Alzheimer\'s disease Diabetes Pneumonia or
influenza Nephropathys Suicide. The period of communicability is less for the communicable
diseases, hence they are prevention of disease in early stages is possible. Prevention and control
of influenza: Good ventilation of public buildings Encouraging sufferers to cover their faces
with a handkerchief when coughing and sneezing Stay in home at the first sign of influenza
Vaccination against the influenza, may be killed or level attenuated vaccines Prevention of
exposure to influenza strains by hygiene practice Antiviral drugs like amantadine or rimantidine
used Prevention and control of pneumonia: Pneumococcal pneumonia vaccination, PPV 23
AND PCV7. Education of mother to observe the sings of pneumonia. History taking and clinical
examination are very important for management Physical examination for, breath count, chest
indrawing, listen for stridor, wheezing and for severe malnutrition, cyanosis Any signs detected,
they are treated early by antibiotics and other measures Improving the primary medical care
services and development..
9. For some non-Mendelian inheritance patterns, the genotype of an i.pdf
9. For some non-Mendelian inheritance patterns, the genotype of an individual is not sufficient to
predict the phenotype. For the following traits, what else do you need to know to predict the
phenotype?
a) Snail coiling direction
b) Angelman syndrome due to deleted UBE3A gene
Solution
The inheritance mechanisms of both snail coiling and Angelman syndrome involve strong
maternal influences. Snail coiling direction is governed by the genotype of the oocyte, not of the
zygote, so much so that the snail coiling direction can be predicted from the maternal genotype
alone. Similarly, Angelman syndrome (due to deletions in ube3A) is governed by the maternal
copy of the gene alone. If the copy of the gene derived from the mother is functional, the
individuals are healthy. If the maternal copy is defective or deficient, the individual suffers from
the Angelman syndrome..
You have mated a phenotypically wild-type female of the genotype Xy+.pdf
You have mated a phenotypically wild-type female of the genotype Xy+w+Xyw to a male with
the genotype XywY. You counted 2205 total F1 progeny and found 17 flies with gray bodies and
white eyes and 12 flies that had yellow body and red eyes. What is the distance between these
genes controlling body color and eye color on the X chromosome?
rev: 10_27_2012
A. 0.78 map units
B. 1.32 map units
C. 2.63 map units
D. 98.7 map units
Solution
In this problem total progeny is 2205. Among them recombinant individuals are 17 flies with
gray body and white eyes; 12 flies with yellow body and red eyes.
Then % of Recombination = No. Of Recombinant Individuals x 100 / Total progeny
= (17 + 12) x 100 / 2205
= 29 x 100 / 2205
= 1.32 % or 1.32 mU.
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Under Augustus, the empire experienced the Pax Romana, a 200-year period of relative peace and stability. Augustus reformed the military, established efficient administrative systems, and initiated grand construction projects. The empire's borders expanded, encompassing territories from Britain to Egypt and from Spain to the Euphrates. Roman legions, renowned for their discipline and engineering prowess, secured and maintained these vast territories, building roads, fortifications, and cities that facilitated control and integration.
The Roman Empire’s society was hierarchical, with a rigid class system. At the top were the patricians, wealthy elites who held significant political power. Below them were the plebeians, free citizens with limited political influence, and the vast numbers of slaves who formed the backbone of the economy. The family unit was central, governed by the paterfamilias, the male head who held absolute authority.
Culturally, the Romans were eclectic, absorbing and adapting elements from the civilizations they encountered, particularly the Greeks. Roman art, literature, and philosophy reflected this synthesis, creating a rich cultural tapestry. Latin, the Roman language, became the lingua franca of the Western world, influencing numerous modern languages.
Roman architecture and engineering achievements were monumental. They perfected the arch, vault, and dome, constructing enduring structures like the Colosseum, Pantheon, and aqueducts. These engineering marvels not only showcased Roman ingenuity but also served practical purposes, from public entertainment to water supply.
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- 1. Genetics Homework Problem 1: You are mapping genes in a species of fruit fly. You cross a wild type fly to a fly from a true breeding mutant strain with green eyes (g), short wings (s w), and hairy legs (hl). The F1 progeny all appear wild type (red eyes, long wings, smooth legs). You then testcross a female F1 fly to a male from the mutant strain, and obtain 200 progeny a. What are the genotypes of the two tiles in the test cross?_____ times ______ b. How many progeny of each phenotype would be expected if the genes were assorting independently? c. Can you reject independent assortment at P=0.05? Use statistical evidence to back up your claim. d. Assuming these genes are all linked on the same chromosome, draw a genetic map showing the order and map distances between the three genes. Solution Answer: a). The genotypes of the two flies in the test cross: Gg Swsw Hlhl X gg swsw hlhl b). If the genes were assorting independently, 8 types of phenotypes are expected in equl proportions 1:1:1:1:1:1:1:1 Green, short, hair: 1 Red, long, smooth: 1 Green, long, smooth: 1 Green, short, smooth: 1 Red, short, hairy: 1 Red, long, hairy: 1 Green, long, hairy: 1 Red, short, hairy: 1 c). Phenotype Observed(O) Expected (E) O-E (O-E)2 (O-E)2/E Green, short, hair: 44 25
- 2. 19 361.00 14.44 Red, long, smooth 42 25 17 289.00 11.56 Green, long, smooth 20 25 -5 25.00 1.00 Green, short, smooth 5 25 -20 400.00 16.00 Red, short, hairy 20 25 -5 25.00 1.00 Red, long, hairy 7 25 -18 324.00 12.96 Green, long, hairy 30 25
- 3. 5 25.00 1.00 Red, short, hairy 32 25 7 49.00 1.96 200 200 59.92 Chi-square value = 59.92 The independent assortment at P=0.05 is rejected. d). Explanation Shorthand Number of progeny g sw hl 44 G Sw Hl 42 g Sw Hl 20 g sw Hl 5 G sw hl 20 G Sw hl 7 g Sw hl 30 G sw Hl 32 Imagine the order is G SW Hl / g sw hl
- 4. 1. If single cross over (SCO) occurs between G & Sw Normal order= G---------Sw & g-----sw After cross over= G-----sw & g------Sw G-----sw recombinants are 20+ 32 = 52 g----- Sw recombinants are 20+ 30 = 50 Total recombinants = 102 RF = (102/200.)*100 =51% 2. If single cross over (SCO) occurs between Sw & Hl Normal order= Sw---------Hl & sw----hl After cross over= Sw-----hl & sw------Hl Sw-----hl recombinants are 30 + 7 = 37 sw------Hl recombinants are 32+5 = 37 Total recombinants = 74 RF = (74/200)*100 = 37% 3. If single cross over (SCO) occurs between G & Hl Normal order= G---------Hl & g------hl After cross over= G-----hl & g------Hl G-----hl recombinants are 20 + 7=27 g------Hl recombinants are 20+5=25 Total recombinants = 52 RF = (52/200)*100 = 26% % RF = Map unit distance The order of genes is ----- G------26 m.u.--------Hl------37 m.u.---------Sw Gene for color of eyes and length of wings are not linked. The gene for legs is linked with gene for wings & gene for eye. Phenotype Observed(O) Expected (E) O-E (O-E)2 (O-E)2/E Green, short, hair: 44 25 19
- 5. 361.00 14.44 Red, long, smooth 42 25 17 289.00 11.56 Green, long, smooth 20 25 -5 25.00 1.00 Green, short, smooth 5 25 -20 400.00 16.00 Red, short, hairy 20 25 -5 25.00 1.00 Red, long, hairy 7 25 -18 324.00 12.96 Green, long, hairy 30 25 5