a. From the DNA fragmentation analysis, it is clear that the polydactyly came from the parents X
chromosome as the affected sons have same band pattern as father has one band that found in
sons and mother has two bands that appeared in affected sons.
The father genotype is XbY, where Xb is affected with polydactyly.
The mother genotype is XbX, one X is unaffected, as daughters are not affected by the disease.
First son has XY genotype, Second son has XbY genotype, First daughter has XbX genotype,
third son has XbY genotype, second daughter has XbX genotype, fourth son has XbY genotype
and fifth son has XY genotype.
b. Yes. For example, AABbccDd genotype might contribute +1.3 units to genetic liability while
aabbCcdd might contribute -2.3 units to genetic liability. Due to unmeasure genetic liability, any
arbitrary unit can be set. It can be 0 which denote the mean liability and positive numbers denote
risk and negative denote protection i.e., lower probablity of developing the disease. Polydactyly
mutation was mapped to chromosome 2p. A single point mutation is maintained to carry the
mutation. A study showed that five-generation Indian family of 37 individuals have 15 affected
persons of polydactyly and genome wide mapping and haplotype analysis observed in this study
indicated highly informative polymorphic markers on part of pedigree showed linkage between
the phenotype and markers on the chromosome.
c. The first daughter is likely to be carrier of this mutation as analyzed from the restriction
fragments of the gene.
Recombination between marker and disease can never be completely ruled out, even for very
tightly linked markers, but the error rate can be greatly reduced by using two marker loci situated
on opposite sides of the disease locus. This type of flanking or bridging markers help to produce
a marker-marker recombination and at least false prediction can be avoided.
Solution
a. From the DNA fragmentation analysis, it is clear that the polydactyly came from the parents X
chromosome as the affected sons have same band pattern as father has one band that found in
sons and mother has two bands that appeared in affected sons.
The father genotype is XbY, where Xb is affected with polydactyly.
The mother genotype is XbX, one X is unaffected, as daughters are not affected by the disease.
First son has XY genotype, Second son has XbY genotype, First daughter has XbX genotype,
third son has XbY genotype, second daughter has XbX genotype, fourth son has XbY genotype
and fifth son has XY genotype.
b. Yes. For example, AABbccDd genotype might contribute +1.3 units to genetic liability while
aabbCcdd might contribute -2.3 units to genetic liability. Due to unmeasure genetic liability, any
arbitrary unit can be set. It can be 0 which denote the mean liability and positive numbers denote
risk and negative denote protection i.e., lower probablity of developing the disease. Polydactyly
mutation was mapped to chromosome 2p. A single point mutation is.
Answer part a)According to the gene map of the wild type and the m.pdfappsmobileshoppe
Answer part a)
According to the gene map of the wild type and the mutant gene, it can be seen that the mutant
lacks a site for restriction digestion. This shows that in wild type, the restriction enzyme can
digest the DNA at any three sites and thus atleast two bands will appear after amplification. On
the contrary, the deletion of middle sequence of restriction site will lead to formation of a single
band in mutant gene. The smaller band will represnt the mutant and the larger band will
represent the normal/wild type allele. Thus, according to this information, it can be deciphered
that the mother is a heterozygous in nature and carries a mutant allele for Tay-Sachs disease
while the father does not. Thus, the lower band will be a representative of mutant or \"t\" and the
upper band will be a representative of normal \"T\" allele.
Answer part b)
Since the mother is a carrier, there are 50% chances that the mother will transmit the mutant
allele to her offspring. Thus, the couple can bear a normal heterozygous carrier female offspring
or an affected male offspring. According to the information in the image, the genotype bands of
father and offspring match which clearly suggest that the offspring is an unaffected heterozygous
carrier female.
Answer part c)
There are always 50% changes for a male to be affected by the disease. Rest 50% males will be
normal both geotypically and phenotypically. Also, all the females will be phenotypically normal
but 50% of them would be heterozygous carriers.
Genotypes: XtX : XtY : XX : XY :: 1:1:1:1
Phenotypes: Affected : normal : carriers :: 1: 2: 1XYXtXtXXtYXXXXY
Solution
Answer part a)
According to the gene map of the wild type and the mutant gene, it can be seen that the mutant
lacks a site for restriction digestion. This shows that in wild type, the restriction enzyme can
digest the DNA at any three sites and thus atleast two bands will appear after amplification. On
the contrary, the deletion of middle sequence of restriction site will lead to formation of a single
band in mutant gene. The smaller band will represnt the mutant and the larger band will
represent the normal/wild type allele. Thus, according to this information, it can be deciphered
that the mother is a heterozygous in nature and carries a mutant allele for Tay-Sachs disease
while the father does not. Thus, the lower band will be a representative of mutant or \"t\" and the
upper band will be a representative of normal \"T\" allele.
Answer part b)
Since the mother is a carrier, there are 50% chances that the mother will transmit the mutant
allele to her offspring. Thus, the couple can bear a normal heterozygous carrier female offspring
or an affected male offspring. According to the information in the image, the genotype bands of
father and offspring match which clearly suggest that the offspring is an unaffected heterozygous
carrier female.
Answer part c)
There are always 50% changes for a male to be affected by the disease. Rest 50% males w.
Linkage refers to the presence of two different genes on the same chromosome . Two genes that occur on the same chromosome are said to be linked, and those that occur very close together are tightly linked.
One case of paternity testing has been given in the figure. Four aut.pdfaashwini4
One case of paternity testing has been given in the figure. Four autoradiographs (autorads) that
show DNA \"fingerprints\" for three individuals: a mother, her child, and the child\'s alleged
father(said father/without proof). Each autorad compares the traits inherited by these three
individuals - four traits are looked at, one for each autorad.
Presence or absence of different alleles (called DNA fingerprints) has been compared in each
case. These alleles (DNA fragments) differ in length from person to person; for this reason they
are used as genetic markers.
In case I-these alleles are A, B, C and D
In case II-E, G and F
In case III-H, I, J and K
In case IV-L, M, N and O
Ans 1
Case
Genotype of Mother
Genotype of Child
Genotype of alleged father
I
A/D
A/C
B/C
II
G/G
E/G
E/F
III
I/J
I/K
H/K
IV
M/N
M/O
L/O
Ans 2
In each case, child has inherited one copy of allele from mother and one from father so by
looking at the genotype of child, contribution from the parents can be predicted (by comparing
the genotype of all three persons)
Case
Genotype of child
Allele inherited from mother
Allele inherited from father
I
A/C
A
C
II
E/G
G
E
III
I/K
I
K
IV
M/O
M
O
Ans-3
In each case, we can see that child has inherited one copy of allele from mother and one from
father. So by making the comparison, it can be predicted that that the alleged father could be the
biological father of this child.
Ans 4
The DNA fingerprint assay is also useful in chimerism analysis after liver transplantation. Four
percent of donor lymphocytes (transient lymphocytes) are present for up to 3 weeks after
transplantation. If the donor lymphocytes persist, they can cause graft-versus-host disease
(GVHD), which is an underdiagnosed and often fatal complication that occurs in approximately
1% of cases, usually 2 to 6 weeks after transplantation. The following cases illustrate the utility
of the this assay in chimerism analysis.
A liver transplant recipient presented with a skin rash 2 weeks after transplant, and a punch
biopsy of the skin was obtained. The pathologist noted the presence of a lymphocyte infiltrate in
the dermis. Possible causes included a drug reaction, a viral infection, or early GVHD. DNA was
extracted from the area of lymphocyte infiltration, and DNA fingerprinting analysis was
performed. The results showed that some of the lymphocytes were of donor origin, confirming
the diagnosis of early GVHD.
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC1200713/
Case
Genotype of Mother
Genotype of Child
Genotype of alleged father
I
A/D
A/C
B/C
II
G/G
E/G
E/F
III
I/J
I/K
H/K
IV
M/N
M/O
L/O
Solution
One case of paternity testing has been given in the figure. Four autoradiographs (autorads) that
show DNA \"fingerprints\" for three individuals: a mother, her child, and the child\'s alleged
father(said father/without proof). Each autorad compares the traits inherited by these three
individuals - four traits are looked at, one for each autorad.
Presence or absence of different alleles.
Answer part a)According to the gene map of the wild type and the m.pdfappsmobileshoppe
Answer part a)
According to the gene map of the wild type and the mutant gene, it can be seen that the mutant
lacks a site for restriction digestion. This shows that in wild type, the restriction enzyme can
digest the DNA at any three sites and thus atleast two bands will appear after amplification. On
the contrary, the deletion of middle sequence of restriction site will lead to formation of a single
band in mutant gene. The smaller band will represnt the mutant and the larger band will
represent the normal/wild type allele. Thus, according to this information, it can be deciphered
that the mother is a heterozygous in nature and carries a mutant allele for Tay-Sachs disease
while the father does not. Thus, the lower band will be a representative of mutant or \"t\" and the
upper band will be a representative of normal \"T\" allele.
Answer part b)
Since the mother is a carrier, there are 50% chances that the mother will transmit the mutant
allele to her offspring. Thus, the couple can bear a normal heterozygous carrier female offspring
or an affected male offspring. According to the information in the image, the genotype bands of
father and offspring match which clearly suggest that the offspring is an unaffected heterozygous
carrier female.
Answer part c)
There are always 50% changes for a male to be affected by the disease. Rest 50% males will be
normal both geotypically and phenotypically. Also, all the females will be phenotypically normal
but 50% of them would be heterozygous carriers.
Genotypes: XtX : XtY : XX : XY :: 1:1:1:1
Phenotypes: Affected : normal : carriers :: 1: 2: 1XYXtXtXXtYXXXXY
Solution
Answer part a)
According to the gene map of the wild type and the mutant gene, it can be seen that the mutant
lacks a site for restriction digestion. This shows that in wild type, the restriction enzyme can
digest the DNA at any three sites and thus atleast two bands will appear after amplification. On
the contrary, the deletion of middle sequence of restriction site will lead to formation of a single
band in mutant gene. The smaller band will represnt the mutant and the larger band will
represent the normal/wild type allele. Thus, according to this information, it can be deciphered
that the mother is a heterozygous in nature and carries a mutant allele for Tay-Sachs disease
while the father does not. Thus, the lower band will be a representative of mutant or \"t\" and the
upper band will be a representative of normal \"T\" allele.
Answer part b)
Since the mother is a carrier, there are 50% chances that the mother will transmit the mutant
allele to her offspring. Thus, the couple can bear a normal heterozygous carrier female offspring
or an affected male offspring. According to the information in the image, the genotype bands of
father and offspring match which clearly suggest that the offspring is an unaffected heterozygous
carrier female.
Answer part c)
There are always 50% changes for a male to be affected by the disease. Rest 50% males w.
Linkage refers to the presence of two different genes on the same chromosome . Two genes that occur on the same chromosome are said to be linked, and those that occur very close together are tightly linked.
One case of paternity testing has been given in the figure. Four aut.pdfaashwini4
One case of paternity testing has been given in the figure. Four autoradiographs (autorads) that
show DNA \"fingerprints\" for three individuals: a mother, her child, and the child\'s alleged
father(said father/without proof). Each autorad compares the traits inherited by these three
individuals - four traits are looked at, one for each autorad.
Presence or absence of different alleles (called DNA fingerprints) has been compared in each
case. These alleles (DNA fragments) differ in length from person to person; for this reason they
are used as genetic markers.
In case I-these alleles are A, B, C and D
In case II-E, G and F
In case III-H, I, J and K
In case IV-L, M, N and O
Ans 1
Case
Genotype of Mother
Genotype of Child
Genotype of alleged father
I
A/D
A/C
B/C
II
G/G
E/G
E/F
III
I/J
I/K
H/K
IV
M/N
M/O
L/O
Ans 2
In each case, child has inherited one copy of allele from mother and one from father so by
looking at the genotype of child, contribution from the parents can be predicted (by comparing
the genotype of all three persons)
Case
Genotype of child
Allele inherited from mother
Allele inherited from father
I
A/C
A
C
II
E/G
G
E
III
I/K
I
K
IV
M/O
M
O
Ans-3
In each case, we can see that child has inherited one copy of allele from mother and one from
father. So by making the comparison, it can be predicted that that the alleged father could be the
biological father of this child.
Ans 4
The DNA fingerprint assay is also useful in chimerism analysis after liver transplantation. Four
percent of donor lymphocytes (transient lymphocytes) are present for up to 3 weeks after
transplantation. If the donor lymphocytes persist, they can cause graft-versus-host disease
(GVHD), which is an underdiagnosed and often fatal complication that occurs in approximately
1% of cases, usually 2 to 6 weeks after transplantation. The following cases illustrate the utility
of the this assay in chimerism analysis.
A liver transplant recipient presented with a skin rash 2 weeks after transplant, and a punch
biopsy of the skin was obtained. The pathologist noted the presence of a lymphocyte infiltrate in
the dermis. Possible causes included a drug reaction, a viral infection, or early GVHD. DNA was
extracted from the area of lymphocyte infiltration, and DNA fingerprinting analysis was
performed. The results showed that some of the lymphocytes were of donor origin, confirming
the diagnosis of early GVHD.
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC1200713/
Case
Genotype of Mother
Genotype of Child
Genotype of alleged father
I
A/D
A/C
B/C
II
G/G
E/G
E/F
III
I/J
I/K
H/K
IV
M/N
M/O
L/O
Solution
One case of paternity testing has been given in the figure. Four autoradiographs (autorads) that
show DNA \"fingerprints\" for three individuals: a mother, her child, and the child\'s alleged
father(said father/without proof). Each autorad compares the traits inherited by these three
individuals - four traits are looked at, one for each autorad.
Presence or absence of different alleles.
2 3 4 5 Assignments -#1 Distinguish between the following genetic term.docxduane12345678
2
3
4
5
Assignments \#1 Distinguish between the following genetic terms: A. Gene and allele. B. Homozygous and heterozygous. C. Genotype and phenotype. D. Dominant and recessive. 1. A man with red-green colour blindness is married to a girl with normal colour vision. What is the probability that his children will be colour blind? 2. A boy has Duchenne muscular dystrophy. His maternal uncle had died of muscular dystrophy at the age of 21 years. What is the probability that this child's sibs will be affected? Explain, giving reasons, whether the following pedigrees are compatible with autosomal dominant, autosomal recessive or X-linked dominant and X -linked recessive inheritance. (Note that a pedigree may be compatible with more than one type of inheritance.) a. b. c. a. b. c. Assignments \#4 Examine the pedigree from a family with a genetic disease and ansv the questions below: a. Does this pedigree indicate autosomal dominant, recessive or sex-link type of inheritance? Give reasons for your choice. b. Assuming that B and b are the normal and mutant alleles respectively what would be the genotypes of the individuals: II.I, II.2 and III.3? c. Individual II.3 requested genetic counselling. What is the probability that her child would be affected. Explain why. Susan's grandfather was deaf, and passed down a hereditary form of deafness within Susan's family as shown in Figure. A. Is this mutation most likely to be dominant or recessive? B. Is it carried on an autosome or a sex chromosome? Why? C. A complete SNP analysis has been done for all of the 11 grandchildren (4 affected, and 7 unaffected). In comparing these 11 SNP results, how long a haplotype block would you expect to find around the critical gene? How might you detect it?
.
Breaking down Biology into simpler bits is the most effective way to learn hence this presentation aims to simplify the concept of 'Linked Inheritance' which makes understanding Inheritance better.
The three components of a nucleotide ... Sugar -.pdfLAMJM
The three components of a nucleotide ... Sugar - Either ribose or deoxyribose (in
RNA or DNA) Nitrogenous base - Adenine, guanine, cytosine, thymine (only in DNA), uracil
(only in RNA) Phosphates - One to three, two of which are usually hydrolyzed to provide the
energy to attach the nucleotide and form the phosphodiester bond
Solution
The three components of a nucleotide ... Sugar - Either ribose or deoxyribose (in
RNA or DNA) Nitrogenous base - Adenine, guanine, cytosine, thymine (only in DNA), uracil
(only in RNA) Phosphates - One to three, two of which are usually hydrolyzed to provide the
energy to attach the nucleotide and form the phosphodiester bond.
More Related Content
Similar to a. From the DNA fragmentation analysis, it is clear that the polydac.pdf
2 3 4 5 Assignments -#1 Distinguish between the following genetic term.docxduane12345678
2
3
4
5
Assignments \#1 Distinguish between the following genetic terms: A. Gene and allele. B. Homozygous and heterozygous. C. Genotype and phenotype. D. Dominant and recessive. 1. A man with red-green colour blindness is married to a girl with normal colour vision. What is the probability that his children will be colour blind? 2. A boy has Duchenne muscular dystrophy. His maternal uncle had died of muscular dystrophy at the age of 21 years. What is the probability that this child's sibs will be affected? Explain, giving reasons, whether the following pedigrees are compatible with autosomal dominant, autosomal recessive or X-linked dominant and X -linked recessive inheritance. (Note that a pedigree may be compatible with more than one type of inheritance.) a. b. c. a. b. c. Assignments \#4 Examine the pedigree from a family with a genetic disease and ansv the questions below: a. Does this pedigree indicate autosomal dominant, recessive or sex-link type of inheritance? Give reasons for your choice. b. Assuming that B and b are the normal and mutant alleles respectively what would be the genotypes of the individuals: II.I, II.2 and III.3? c. Individual II.3 requested genetic counselling. What is the probability that her child would be affected. Explain why. Susan's grandfather was deaf, and passed down a hereditary form of deafness within Susan's family as shown in Figure. A. Is this mutation most likely to be dominant or recessive? B. Is it carried on an autosome or a sex chromosome? Why? C. A complete SNP analysis has been done for all of the 11 grandchildren (4 affected, and 7 unaffected). In comparing these 11 SNP results, how long a haplotype block would you expect to find around the critical gene? How might you detect it?
.
Breaking down Biology into simpler bits is the most effective way to learn hence this presentation aims to simplify the concept of 'Linked Inheritance' which makes understanding Inheritance better.
The three components of a nucleotide ... Sugar -.pdfLAMJM
The three components of a nucleotide ... Sugar - Either ribose or deoxyribose (in
RNA or DNA) Nitrogenous base - Adenine, guanine, cytosine, thymine (only in DNA), uracil
(only in RNA) Phosphates - One to three, two of which are usually hydrolyzed to provide the
energy to attach the nucleotide and form the phosphodiester bond
Solution
The three components of a nucleotide ... Sugar - Either ribose or deoxyribose (in
RNA or DNA) Nitrogenous base - Adenine, guanine, cytosine, thymine (only in DNA), uracil
(only in RNA) Phosphates - One to three, two of which are usually hydrolyzed to provide the
energy to attach the nucleotide and form the phosphodiester bond.
The alcohol is a secondary alcohol,therefore must.pdfLAMJM
The alcohol is a secondary alcohol,therefore must have been made from reduction
(using metalhydrides) of ketones (It\'s not letting me insert diagrams, but the structure is1-
phenyl-2-propanone)
Solution
The alcohol is a secondary alcohol,therefore must have been made from reduction
(using metalhydrides) of ketones (It\'s not letting me insert diagrams, but the structure is1-
phenyl-2-propanone).
rate = K[R1][R2] Order of the reaction is the sum.pdfLAMJM
rate = K[R1][R2] Order of the reaction is the sum of the powers of the
concentration terms of the reactants in the rate expression. So order = 1+1 = 2 ---> 2nd order
reaction
Solution
rate = K[R1][R2] Order of the reaction is the sum of the powers of the
concentration terms of the reactants in the rate expression. So order = 1+1 = 2 ---> 2nd order
reaction.
moles of Fe = 1.2656 = 0.0225 moles of O2 = (1.8.pdfLAMJM
moles of Fe = 1.26/56 = 0.0225 moles of O2 = (1.8-1.26)/16 = 0.03375 so, for
every atom of Fe there is 3/2 moles of O so, empirical formula is Fe2O3
Solution
moles of Fe = 1.26/56 = 0.0225 moles of O2 = (1.8-1.26)/16 = 0.03375 so, for
every atom of Fe there is 3/2 moles of O so, empirical formula is Fe2O3.
In SO2, the charge of O is 2(-2)= -4 so the charg.pdfLAMJM
In SO2, the charge of O is 2(-2)= -4 so the charge of S is +4. In H2SO4, the charge
of O is 4(-2)= -8, the charge of H is 2(+1)= +2, so the charge of S is +6. Since S +4 becomes S
+6, electrons are lost, and this is oxidation. So, SO2 is the substance oxidized, and it is also the
reducing agent. The other reactant, HNO3, is the substance reduced, and it is the oxidizing agent.
Solution
In SO2, the charge of O is 2(-2)= -4 so the charge of S is +4. In H2SO4, the charge
of O is 4(-2)= -8, the charge of H is 2(+1)= +2, so the charge of S is +6. Since S +4 becomes S
+6, electrons are lost, and this is oxidation. So, SO2 is the substance oxidized, and it is also the
reducing agent. The other reactant, HNO3, is the substance reduced, and it is the oxidizing agent..
Elution order is based on the polarities of the c.pdfLAMJM
Elution order is based on the polarities of the compounds. Remember the saying
\"like dissolves like\" when thinking about theanswer. The mobile phase starts out very non-polar
usinghexanes; so the compound that will elute first is the mostnon-polar compound. The
compound that is more polar than oneof the other compounds but not as polar as the third
compound willelute second, and the most polar compound will elute third. Adding
dichloromethane speeds up the elution of the more polarcompounds. Naphthalene is the most
non-polar compound due tothe fact that it only has carbons and hydrogens. Fluorenol ismostly
carbons and hydrogens, but it contains a hydroxy groupmaking it a little more polar than
naphthalene. o-Toluic acidcontains a carboxylic acid group making it the most polar
compoundof the three. So, the elution order for these three compoundsis: naphthalene, then
fluorenol, and then o-toluic acid.
Solution
Elution order is based on the polarities of the compounds. Remember the saying
\"like dissolves like\" when thinking about theanswer. The mobile phase starts out very non-polar
usinghexanes; so the compound that will elute first is the mostnon-polar compound. The
compound that is more polar than oneof the other compounds but not as polar as the third
compound willelute second, and the most polar compound will elute third. Adding
dichloromethane speeds up the elution of the more polarcompounds. Naphthalene is the most
non-polar compound due tothe fact that it only has carbons and hydrogens. Fluorenol ismostly
carbons and hydrogens, but it contains a hydroxy groupmaking it a little more polar than
naphthalene. o-Toluic acidcontains a carboxylic acid group making it the most polar
compoundof the three. So, the elution order for these three compoundsis: naphthalene, then
fluorenol, and then o-toluic acid..
XSCID - Human Severe Combined Immunodeficiency SyndromeIt is an P.pdfLAMJM
XSCID :- Human Severe Combined Immunodeficiency Syndrome
It is an Primary Immunodeficiency disorder, which caused due to mutation in the gene on the X
Chromosome,that is essential for T cell growth factor(Gamma c) for formation of Gamma Chain,
due to which T cells wont differentiate thus T cells will decrease in count along with NK cells,
but the B cells will be very high,where T cells plays an vital role in the eradication of pathogens
from the body,not completed by the B cells due to absence of its receptors, so it causes
dysfunction,imbalance in the system thus causes immunodeficiency.
It is reported in high number due to X linked defective gene, nearly 45% of all cases are reported
as X SCID.
Difference between X SCID and IL2 deficiency :-
Comparision : Both are immuno deficiency, caused due to impaired T cell differentiation, except
few most of the symptoms are quite similar,
Il2 deficiency is due to alpha receptor dysfunction is also called CD25
dysfunction.S.NoFeatureX SCIDIL2 deficiency1EpidemiologyReport more casesRare
cases2SeverityMore severe due to absence of T cells, high B cellsPoor differentiation of T and B
cells3Phenotypic appearancePrimary ImmunodeficiencyResembles like immune syndromes like
Polyendocrinopathy, IPEX4SymptomsSensitive for diseases, no characteristic
locationInflammation of lymphatic tissues like lung,liver,gut bone.
Solution
XSCID :- Human Severe Combined Immunodeficiency Syndrome
It is an Primary Immunodeficiency disorder, which caused due to mutation in the gene on the X
Chromosome,that is essential for T cell growth factor(Gamma c) for formation of Gamma Chain,
due to which T cells wont differentiate thus T cells will decrease in count along with NK cells,
but the B cells will be very high,where T cells plays an vital role in the eradication of pathogens
from the body,not completed by the B cells due to absence of its receptors, so it causes
dysfunction,imbalance in the system thus causes immunodeficiency.
It is reported in high number due to X linked defective gene, nearly 45% of all cases are reported
as X SCID.
Difference between X SCID and IL2 deficiency :-
Comparision : Both are immuno deficiency, caused due to impaired T cell differentiation, except
few most of the symptoms are quite similar,
Il2 deficiency is due to alpha receptor dysfunction is also called CD25
dysfunction.S.NoFeatureX SCIDIL2 deficiency1EpidemiologyReport more casesRare
cases2SeverityMore severe due to absence of T cells, high B cellsPoor differentiation of T and B
cells3Phenotypic appearancePrimary ImmunodeficiencyResembles like immune syndromes like
Polyendocrinopathy, IPEX4SymptomsSensitive for diseases, no characteristic
locationInflammation of lymphatic tissues like lung,liver,gut bone..
p = 0.25n = 1000standard error, SE = sqrt(p(1-p)n)SE = sq.pdfLAMJM
p = 0.25
n = 1000
standard error, SE = sqrt(p(1-p)/n)
SE = sqrt(0.25(1-0.25) / 1000) = 0.0137
.
proportion of flights within 3%
= P(-0.03/0.0137
= P( (-0.03 / 0.0137)
Solution
p = 0.25
n = 1000
standard error, SE = sqrt(p(1-p)/n)
SE = sqrt(0.25(1-0.25) / 1000) = 0.0137
.
proportion of flights within 3%
= P(-0.03/0.0137
= P( (-0.03 / 0.0137).
CaO is the lewis base (accepts electrons). SO2 is.pdfLAMJM
CaO is the lewis base (accepts electrons). SO2 is the lewis acid (donates electrons).
Solution
CaO is the lewis base (accepts electrons). SO2 is the lewis acid (donates electrons)..
The thrust to weight ratio must be higer than one in order to get (V.pdfLAMJM
The thrust to weight ratio must be higer than one in order to get (VTOL) Vertical lift off and
landing. This ratio must be greater than one to lift from the ground with the thrust vector pointed
downward during take off and pointed back ward during normal flight. In earlier days the main
problem with these aircraft was that thier control was too dificult and the maintenence of engine
was difficult too. In this way accident rate or ratio was too high. SO , Harrier was the first air
craft with successfull VTOL. There were four vectorizing Nozzels which could be turned
vectorially from 0 to 98degrees. WIth the reaction control system with the extra set of small
thruster there were no air stream in hover flight accross any control system . Harrier was prior to
preveous such jets because of its high roll rate, a small turn radius , and its ability of climb to
high altitude abruptly. Another plus point of Horrier was that it lands just like an ordinary plane
vectoring the thrust to about 45 degree down .
Solution
The thrust to weight ratio must be higer than one in order to get (VTOL) Vertical lift off and
landing. This ratio must be greater than one to lift from the ground with the thrust vector pointed
downward during take off and pointed back ward during normal flight. In earlier days the main
problem with these aircraft was that thier control was too dificult and the maintenence of engine
was difficult too. In this way accident rate or ratio was too high. SO , Harrier was the first air
craft with successfull VTOL. There were four vectorizing Nozzels which could be turned
vectorially from 0 to 98degrees. WIth the reaction control system with the extra set of small
thruster there were no air stream in hover flight accross any control system . Harrier was prior to
preveous such jets because of its high roll rate, a small turn radius , and its ability of climb to
high altitude abruptly. Another plus point of Horrier was that it lands just like an ordinary plane
vectoring the thrust to about 45 degree down ..
Views are virtual tables which are not real tables but are generated.pdfLAMJM
Views are virtual tables which are not real tables but are generated from base tables
Solution
Views are virtual tables which are not real tables but are generated from base tables.
This was an observational study as the investigators obsereved the e.pdfLAMJM
This was an observational study as the investigators obsereved the effect of proximity on the
interference of competition between two ants species.
The data that was used were gathered from the earlier studies as well as the research conducted
by the various researchers of the particular institute. They had mapped and censused the data
since the summer of 1985 for the ant population Pogonomyrmex (each of the colony was
labelled individually, and located using coordinates from a fixed origin, using a fixed infrared
theodolite. ), and since 1997 they had mapped and censused every Aphaenogaster member on the
particular site(aggression test was performed).
They studied the nesting pattern.
The age of the colony was percieved by the mating pattern of the ants.
By the cooperation among the habitants they found whether the ants belonged to same colony or
the other.
They saw that there was a significant effect of distance among the colonies,between teo different
species and also among the same species.
Solution
This was an observational study as the investigators obsereved the effect of proximity on the
interference of competition between two ants species.
The data that was used were gathered from the earlier studies as well as the research conducted
by the various researchers of the particular institute. They had mapped and censused the data
since the summer of 1985 for the ant population Pogonomyrmex (each of the colony was
labelled individually, and located using coordinates from a fixed origin, using a fixed infrared
theodolite. ), and since 1997 they had mapped and censused every Aphaenogaster member on the
particular site(aggression test was performed).
They studied the nesting pattern.
The age of the colony was percieved by the mating pattern of the ants.
By the cooperation among the habitants they found whether the ants belonged to same colony or
the other.
They saw that there was a significant effect of distance among the colonies,between teo different
species and also among the same species..
System integrationits defined as the development of carrying organ.pdfLAMJM
System integration
its defined as the development of carrying organized the element subsystems single system and
certifying the subsystems role organized as a system. In evidence skill, systems integration is the
process of connecting composed changed computing systems and software submissions
substantially or functionally to performance as a corresponding complete
System integration (SI) is as well about calculation value the system, competences that are
conceivable since of connections among subsystems
A system integration cause wants a wide range of services and is possible to be clear by a extent
of information relatively than a complexity of information. These services are likely to contain
software, systems then initiative manner, software and hardware engineering, border, and overall
problematic resolving skills. It is expected that the hitches to be resolved have not stayed solved
earlier but in the widest sense. They are likely to comprise new and interesting complications
with an effort from a comprehensive variety of causes wherever the system integration cause
Vertical integration
Vertical integration is the course of mixing subsystems conferring to their functionality by
making efficient objects also mentioned to as storage tower. The advantage of this technique is
that the integration is done rapidly and includes only the essential sellers, so, this technique is
inexpensive in the small term.
Star integration
It’s also known as spaghetti integration, is a procedure of schemes mixing anywhere to each
system is interrelated to apiece of the residual subsystems. Once detected since the belvedere of
the subsystem which is lifecycle joint, the systems are suggestive of a star, but after the general
diagram of the scheme is accessible, the influences look like spaghetti, from now the title of this
process.
Horizontal integration
Horizontal integration is an mixing way in which a dedicated subsystem is devoted to
announcement among further subsystems. This agrees critical the amount of influences first
unique per subsystem which drive link directly to the ESB. The ESB is skilled of decoding the
border into extra interface. This lets wounding the costs of addition and affords dangerous
suppleness.
Internal ownership
On the other pointer, cost-of-ownership can be significantly advanced than seen in other devices,
since in situation of original or improved functionality, the only conceivable way to tool would
be by applying added silo. Recycling subsystems to generate one more functionality is not
possible.
External (third parties) ownership
A third party is an separate or thing that is complex in a deal then is not one of the leaders then
has a smaller attention. Ownership is most simply established in the case of perceptible property.
But, in the case of intelligent stuff which is critical for software. Most similes of intelligent
possessions favor it as analogous to, then not the identical as, palpable stuff
Solution
System integration.
Solution Function of neuron is to transmit information to the oth.pdfLAMJM
Solution
:
Function of neuron is to transmit information to the other neuron after processing it.It generates
nerve implulses.Neurones play their role or transmit information through electrical nerve
impulses.These electrical nerve information are generated by the action of some chemicals or we
can say that nerve impulses use electric power generated by chemicals for transmission of
information..
June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...Levi Shapiro
Letter from the Congress of the United States regarding Anti-Semitism sent June 3rd to MIT President Sally Kornbluth, MIT Corp Chair, Mark Gorenberg
Dear Dr. Kornbluth and Mr. Gorenberg,
The US House of Representatives is deeply concerned by ongoing and pervasive acts of antisemitic
harassment and intimidation at the Massachusetts Institute of Technology (MIT). Failing to act decisively to ensure a safe learning environment for all students would be a grave dereliction of your responsibilities as President of MIT and Chair of the MIT Corporation.
This Congress will not stand idly by and allow an environment hostile to Jewish students to persist. The House believes that your institution is in violation of Title VI of the Civil Rights Act, and the inability or
unwillingness to rectify this violation through action requires accountability.
Postsecondary education is a unique opportunity for students to learn and have their ideas and beliefs challenged. However, universities receiving hundreds of millions of federal funds annually have denied
students that opportunity and have been hijacked to become venues for the promotion of terrorism, antisemitic harassment and intimidation, unlawful encampments, and in some cases, assaults and riots.
The House of Representatives will not countenance the use of federal funds to indoctrinate students into hateful, antisemitic, anti-American supporters of terrorism. Investigations into campus antisemitism by the Committee on Education and the Workforce and the Committee on Ways and Means have been expanded into a Congress-wide probe across all relevant jurisdictions to address this national crisis. The undersigned Committees will conduct oversight into the use of federal funds at MIT and its learning environment under authorities granted to each Committee.
• The Committee on Education and the Workforce has been investigating your institution since December 7, 2023. The Committee has broad jurisdiction over postsecondary education, including its compliance with Title VI of the Civil Rights Act, campus safety concerns over disruptions to the learning environment, and the awarding of federal student aid under the Higher Education Act.
• The Committee on Oversight and Accountability is investigating the sources of funding and other support flowing to groups espousing pro-Hamas propaganda and engaged in antisemitic harassment and intimidation of students. The Committee on Oversight and Accountability is the principal oversight committee of the US House of Representatives and has broad authority to investigate “any matter” at “any time” under House Rule X.
• The Committee on Ways and Means has been investigating several universities since November 15, 2023, when the Committee held a hearing entitled From Ivory Towers to Dark Corners: Investigating the Nexus Between Antisemitism, Tax-Exempt Universities, and Terror Financing. The Committee followed the hearing with letters to those institutions on January 10, 202
Synthetic Fiber Construction in lab .pptxPavel ( NSTU)
Synthetic fiber production is a fascinating and complex field that blends chemistry, engineering, and environmental science. By understanding these aspects, students can gain a comprehensive view of synthetic fiber production, its impact on society and the environment, and the potential for future innovations. Synthetic fibers play a crucial role in modern society, impacting various aspects of daily life, industry, and the environment. ynthetic fibers are integral to modern life, offering a range of benefits from cost-effectiveness and versatility to innovative applications and performance characteristics. While they pose environmental challenges, ongoing research and development aim to create more sustainable and eco-friendly alternatives. Understanding the importance of synthetic fibers helps in appreciating their role in the economy, industry, and daily life, while also emphasizing the need for sustainable practices and innovation.
Acetabularia Information For Class 9 .docxvaibhavrinwa19
Acetabularia acetabulum is a single-celled green alga that in its vegetative state is morphologically differentiated into a basal rhizoid and an axially elongated stalk, which bears whorls of branching hairs. The single diploid nucleus resides in the rhizoid.
Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
Model Attribute Check Company Auto PropertyCeline George
In Odoo, the multi-company feature allows you to manage multiple companies within a single Odoo database instance. Each company can have its own configurations while still sharing common resources such as products, customers, and suppliers.
Honest Reviews of Tim Han LMA Course Program.pptxtimhan337
Personal development courses are widely available today, with each one promising life-changing outcomes. Tim Han’s Life Mastery Achievers (LMA) Course has drawn a lot of interest. In addition to offering my frank assessment of Success Insider’s LMA Course, this piece examines the course’s effects via a variety of Tim Han LMA course reviews and Success Insider comments.
Macroeconomics- Movie Location
This will be used as part of your Personal Professional Portfolio once graded.
Objective:
Prepare a presentation or a paper using research, basic comparative analysis, data organization and application of economic information. You will make an informed assessment of an economic climate outside of the United States to accomplish an entertainment industry objective.
Biological screening of herbal drugs: Introduction and Need for
Phyto-Pharmacological Screening, New Strategies for evaluating
Natural Products, In vitro evaluation techniques for Antioxidants, Antimicrobial and Anticancer drugs. In vivo evaluation techniques
for Anti-inflammatory, Antiulcer, Anticancer, Wound healing, Antidiabetic, Hepatoprotective, Cardio protective, Diuretics and
Antifertility, Toxicity studies as per OECD guidelines
a. From the DNA fragmentation analysis, it is clear that the polydac.pdf
1. a. From the DNA fragmentation analysis, it is clear that the polydactyly came from the parents X
chromosome as the affected sons have same band pattern as father has one band that found in
sons and mother has two bands that appeared in affected sons.
The father genotype is XbY, where Xb is affected with polydactyly.
The mother genotype is XbX, one X is unaffected, as daughters are not affected by the disease.
First son has XY genotype, Second son has XbY genotype, First daughter has XbX genotype,
third son has XbY genotype, second daughter has XbX genotype, fourth son has XbY genotype
and fifth son has XY genotype.
b. Yes. For example, AABbccDd genotype might contribute +1.3 units to genetic liability while
aabbCcdd might contribute -2.3 units to genetic liability. Due to unmeasure genetic liability, any
arbitrary unit can be set. It can be 0 which denote the mean liability and positive numbers denote
risk and negative denote protection i.e., lower probablity of developing the disease. Polydactyly
mutation was mapped to chromosome 2p. A single point mutation is maintained to carry the
mutation. A study showed that five-generation Indian family of 37 individuals have 15 affected
persons of polydactyly and genome wide mapping and haplotype analysis observed in this study
indicated highly informative polymorphic markers on part of pedigree showed linkage between
the phenotype and markers on the chromosome.
c. The first daughter is likely to be carrier of this mutation as analyzed from the restriction
fragments of the gene.
Recombination between marker and disease can never be completely ruled out, even for very
tightly linked markers, but the error rate can be greatly reduced by using two marker loci situated
on opposite sides of the disease locus. This type of flanking or bridging markers help to produce
a marker-marker recombination and at least false prediction can be avoided.
Solution
a. From the DNA fragmentation analysis, it is clear that the polydactyly came from the parents X
chromosome as the affected sons have same band pattern as father has one band that found in
sons and mother has two bands that appeared in affected sons.
The father genotype is XbY, where Xb is affected with polydactyly.
The mother genotype is XbX, one X is unaffected, as daughters are not affected by the disease.
First son has XY genotype, Second son has XbY genotype, First daughter has XbX genotype,
third son has XbY genotype, second daughter has XbX genotype, fourth son has XbY genotype
and fifth son has XY genotype.
b. Yes. For example, AABbccDd genotype might contribute +1.3 units to genetic liability while
2. aabbCcdd might contribute -2.3 units to genetic liability. Due to unmeasure genetic liability, any
arbitrary unit can be set. It can be 0 which denote the mean liability and positive numbers denote
risk and negative denote protection i.e., lower probablity of developing the disease. Polydactyly
mutation was mapped to chromosome 2p. A single point mutation is maintained to carry the
mutation. A study showed that five-generation Indian family of 37 individuals have 15 affected
persons of polydactyly and genome wide mapping and haplotype analysis observed in this study
indicated highly informative polymorphic markers on part of pedigree showed linkage between
the phenotype and markers on the chromosome.
c. The first daughter is likely to be carrier of this mutation as analyzed from the restriction
fragments of the gene.
Recombination between marker and disease can never be completely ruled out, even for very
tightly linked markers, but the error rate can be greatly reduced by using two marker loci situated
on opposite sides of the disease locus. This type of flanking or bridging markers help to produce
a marker-marker recombination and at least false prediction can be avoided.
