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.
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.
With X-linked inheritance, the X chromosome is transmitted in a diffe.pdfarihantstoneart
With X-linked inheritance, the X chromosome is transmitted in a different pattern by males and
females. In X-linked dominant traits, males hemizygous for trait will express the trait, but
females heterozygous will be carriers. For X-linked recessive, none of the sons of an affected
female will show the trait. For autosomal dominant, two affected parents may produce an
unaffected child.
Solution
Answer:
A. True
Most X-linked conditions are recessive. This means that in a person with two X chromosomes
(most females), both copies of a gene (i.e., one on each X chromosome) must have a change or
mutation whereas in a person with one X chromosome (most males), only one copy of a gene
must have a mutation. A female with a mutation in one copy of a gene on the X chromosome is
said to be a “carrier” for an X-linked condition. A male with a mutation in a gene on the X
chromosome is typically affected with the condition.
The father\'s X-chromosome goes to the daughter while the mother can give one X-chromosome
to the son and other to the daughter.
B. False
For an X-linked dominant condition, only one copy of a gene on the X chromosome whether in a
female with two X chromosomes or males with on X chromosome must have a change or
mutation for an individual to be affected with the condition. For this reason, X-linked disorders
are often seen with similar frequency in males and females. However, since females also have
one normal X chromosome as well as an X chromosome with a mutation, the condition is often
more “mild.”
C.False
For X-linked recessive disorders, an affected father who has a mutation in a gene on the X
chromosome can transmit either the X chromosome with this mutation or a Y chromosome to his
children. If the mother is not affected or a carrier, none of his sons will be affected since they can
only inherit a normal X chromosome from their mother and they inherit a Y chromosome from
their father.
But if the female is affected, it will receive one of the affected X chromosomes, which is enough
to express the trait in the son.
D. True
A person affected by an autosomal dominant disorder has a 50 percent chance of passing the
mutated gene to each child. The chance that a child will not inherit the mutated gene is also 50
percent..
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.
With X-linked inheritance, the X chromosome is transmitted in a diffe.pdfarihantstoneart
With X-linked inheritance, the X chromosome is transmitted in a different pattern by males and
females. In X-linked dominant traits, males hemizygous for trait will express the trait, but
females heterozygous will be carriers. For X-linked recessive, none of the sons of an affected
female will show the trait. For autosomal dominant, two affected parents may produce an
unaffected child.
Solution
Answer:
A. True
Most X-linked conditions are recessive. This means that in a person with two X chromosomes
(most females), both copies of a gene (i.e., one on each X chromosome) must have a change or
mutation whereas in a person with one X chromosome (most males), only one copy of a gene
must have a mutation. A female with a mutation in one copy of a gene on the X chromosome is
said to be a “carrier” for an X-linked condition. A male with a mutation in a gene on the X
chromosome is typically affected with the condition.
The father\'s X-chromosome goes to the daughter while the mother can give one X-chromosome
to the son and other to the daughter.
B. False
For an X-linked dominant condition, only one copy of a gene on the X chromosome whether in a
female with two X chromosomes or males with on X chromosome must have a change or
mutation for an individual to be affected with the condition. For this reason, X-linked disorders
are often seen with similar frequency in males and females. However, since females also have
one normal X chromosome as well as an X chromosome with a mutation, the condition is often
more “mild.”
C.False
For X-linked recessive disorders, an affected father who has a mutation in a gene on the X
chromosome can transmit either the X chromosome with this mutation or a Y chromosome to his
children. If the mother is not affected or a carrier, none of his sons will be affected since they can
only inherit a normal X chromosome from their mother and they inherit a Y chromosome from
their father.
But if the female is affected, it will receive one of the affected X chromosomes, which is enough
to express the trait in the son.
D. True
A person affected by an autosomal dominant disorder has a 50 percent chance of passing the
mutated gene to each child. The chance that a child will not inherit the mutated gene is also 50
percent..
a) How are X-linked genes inherited in humans Be able to use this i.pdfmampbellzumberge517
a) How are X-linked genes inherited in humans? Be able to use this information to perform
monohybrid and dihybrid crosses and to interpret pedigrees. (SHOW WORK & EXPLAIN)
b) Describe dosage compensation/X-inactivation and its consequences. (SHOW WORK &
EXPLAIN)
c) Describe Y-linked inheritance and be able to use this information in monohybrid and dihybrid
crosses and pedigrees. (SHOW WORK & EXPLAIN)
d) Explain complete dominance, incomplete dominance, codominance, penetrance, expressivity,
lethal alleles and multiple alleles and give examples of each. (SHOW WORK & EXPLAIN)
Solution
If cross is made between XY MALE and Xx carrier female then offspring will as shown below
50% hemophilic Xx ( carrier) female and hemophilic boy Yx
50% normal XX and XY.
2.In many animals male have one and female have two X chromosome . The difference in X
chromosome dosage between the two sexes is compensated by mechanism that regulate X
chromosome transcription. Mainteneance of chromosome dosage is necessary in normal
development. X chromosome contain many gene that are important in both sexes male contain
single X and female contain two X chromosome. .To compensate for X chromosome dosage
difference between sexes different mechanism have evolved to equalize x linked transcript level
in male and female..
polypoidy refers to increase in dosage in all chromosome while aneuploidy refers to change in
dosage of one chromosome with respect to rest of the genome.
Consequence of X gene inactivation is visibly mosaic phenotypes exhibited by mammalian
female heterozygous for X linked genes affecting hair and skin .
Answer 3: Y linked inheritance can be seen in male members as only male has Y chromosome.
Y-linked inheritance is appearance of traits in an organism due to the presence of an alelle on the
Y-chromosome e.g hypertrichosis and pattern baldness are due to Y linked inheritance.
Answer 4; Complete dominance is form of dominance in hetrozygous allele that is dominant
completely mask the effect of allele that is recessive flower on mendel\'s pea plant is example of
complete dominance Blue flower has dominant allele B whle b is recessive cross will be
dominant are BB, Bb,Bb with blue flowers
Incomplete dominance is a form of inheritance in which one allele for a specific trait is not
completely expressed over its paired allele .As a result third phenotype appear which is the
combination of both allele e. g snapdragon flower is pink when cross between red flower and
white flower here neither white is dominant nor red
Codominance is form of dominance where in allele of a gene pair in a heterozygote are fully
expressedXYXXXXYxXxxY.
Can yyou answer number 5 please advance genetic As discussed in clas.pdfvikasbajajhissar
Can yyou answer number 5 please advance genetic As discussed in class, the dark (melanic)
form of the peppered moth, Biston betularia, has a survival advantage in industrialized regions.
The melanic allele is dominant to the typical (light) allele. In a particular forest within an
industrialized region, the frequency of the melanic allele is 0.7 and the typical allele is 0.3. The
light form of moths have a reproductive success that 47% that of the dark form. What will the
allele frequencies be after one generation of selection? Given a population that up to now had
been in Hardy-Weinberg equilibrium. Assume two alleles, one locus, rho = 0.5, and distinctly
different (and unambiguous) phenotypes associated with each genotype. Now assume internal
fertilization and that all matings over one generation are 100% assortative with regard to the trait
in question. What are the genotype frequencies before the round of assortative mating? What are
the genotype frequencies in the generation that follows this round of assortative mating?
Solution
A) The genotype frequencies during the Hardy-Weinberg equlibrium must be favouring
heterozygosity i.e. any individual with any given genotype has the equal chance of mating with
another individual of any given genotype. The equlibrium of individual allele frequencies and the
proportion of the various genotypic combinations is established when p value statistically
significant value is 0.5 the disimilar genotypic populations mates together.
B) Assortive mating leads to an over abundance of homozygous individuals Which are likely to
share similar genotypes as When matings of similar phenotypes occur more frequently than by
random chance, the likelihood of offspring receiving two copies of an identical allele increases,
disrupting the Hardy-Weinberg expectations.
There will be the increase in total population variance as homozygosity will be increased for
example if we have a trait controlled by 2 loci then at equilibrium would deviate one from
another significantly.
Assortative mating can act as a powerful mechanism for genetic change in a population specially
where phenotype can be attributed to the interaction at 2 loci with no dominance, it is easier that
100% assortative mating results in an overall increase in homozygosity and total population
variance thus affecting genotype frequencies..
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.
a. From the DNA fragmentation analysis, it is clear that the polydac.pdfLAMJM
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.
a) How are X-linked genes inherited in humans Be able to use this i.pdfmampbellzumberge517
a) How are X-linked genes inherited in humans? Be able to use this information to perform
monohybrid and dihybrid crosses and to interpret pedigrees. (SHOW WORK & EXPLAIN)
b) Describe dosage compensation/X-inactivation and its consequences. (SHOW WORK &
EXPLAIN)
c) Describe Y-linked inheritance and be able to use this information in monohybrid and dihybrid
crosses and pedigrees. (SHOW WORK & EXPLAIN)
d) Explain complete dominance, incomplete dominance, codominance, penetrance, expressivity,
lethal alleles and multiple alleles and give examples of each. (SHOW WORK & EXPLAIN)
Solution
If cross is made between XY MALE and Xx carrier female then offspring will as shown below
50% hemophilic Xx ( carrier) female and hemophilic boy Yx
50% normal XX and XY.
2.In many animals male have one and female have two X chromosome . The difference in X
chromosome dosage between the two sexes is compensated by mechanism that regulate X
chromosome transcription. Mainteneance of chromosome dosage is necessary in normal
development. X chromosome contain many gene that are important in both sexes male contain
single X and female contain two X chromosome. .To compensate for X chromosome dosage
difference between sexes different mechanism have evolved to equalize x linked transcript level
in male and female..
polypoidy refers to increase in dosage in all chromosome while aneuploidy refers to change in
dosage of one chromosome with respect to rest of the genome.
Consequence of X gene inactivation is visibly mosaic phenotypes exhibited by mammalian
female heterozygous for X linked genes affecting hair and skin .
Answer 3: Y linked inheritance can be seen in male members as only male has Y chromosome.
Y-linked inheritance is appearance of traits in an organism due to the presence of an alelle on the
Y-chromosome e.g hypertrichosis and pattern baldness are due to Y linked inheritance.
Answer 4; Complete dominance is form of dominance in hetrozygous allele that is dominant
completely mask the effect of allele that is recessive flower on mendel\'s pea plant is example of
complete dominance Blue flower has dominant allele B whle b is recessive cross will be
dominant are BB, Bb,Bb with blue flowers
Incomplete dominance is a form of inheritance in which one allele for a specific trait is not
completely expressed over its paired allele .As a result third phenotype appear which is the
combination of both allele e. g snapdragon flower is pink when cross between red flower and
white flower here neither white is dominant nor red
Codominance is form of dominance where in allele of a gene pair in a heterozygote are fully
expressedXYXXXXYxXxxY.
Can yyou answer number 5 please advance genetic As discussed in clas.pdfvikasbajajhissar
Can yyou answer number 5 please advance genetic As discussed in class, the dark (melanic)
form of the peppered moth, Biston betularia, has a survival advantage in industrialized regions.
The melanic allele is dominant to the typical (light) allele. In a particular forest within an
industrialized region, the frequency of the melanic allele is 0.7 and the typical allele is 0.3. The
light form of moths have a reproductive success that 47% that of the dark form. What will the
allele frequencies be after one generation of selection? Given a population that up to now had
been in Hardy-Weinberg equilibrium. Assume two alleles, one locus, rho = 0.5, and distinctly
different (and unambiguous) phenotypes associated with each genotype. Now assume internal
fertilization and that all matings over one generation are 100% assortative with regard to the trait
in question. What are the genotype frequencies before the round of assortative mating? What are
the genotype frequencies in the generation that follows this round of assortative mating?
Solution
A) The genotype frequencies during the Hardy-Weinberg equlibrium must be favouring
heterozygosity i.e. any individual with any given genotype has the equal chance of mating with
another individual of any given genotype. The equlibrium of individual allele frequencies and the
proportion of the various genotypic combinations is established when p value statistically
significant value is 0.5 the disimilar genotypic populations mates together.
B) Assortive mating leads to an over abundance of homozygous individuals Which are likely to
share similar genotypes as When matings of similar phenotypes occur more frequently than by
random chance, the likelihood of offspring receiving two copies of an identical allele increases,
disrupting the Hardy-Weinberg expectations.
There will be the increase in total population variance as homozygosity will be increased for
example if we have a trait controlled by 2 loci then at equilibrium would deviate one from
another significantly.
Assortative mating can act as a powerful mechanism for genetic change in a population specially
where phenotype can be attributed to the interaction at 2 loci with no dominance, it is easier that
100% assortative mating results in an overall increase in homozygosity and total population
variance thus affecting genotype frequencies..
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.
a. From the DNA fragmentation analysis, it is clear that the polydac.pdfLAMJM
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.
The name of As2S5 is diarsenic pentasulfide.TRUEExplanationAs .pdfappsmobileshoppe
The name of As2S5 is diarsenic pentasulfide.
TRUE
Explanation
As there are 2 atoms of Arsenic so we uise word\"di\"
As there are 5 atoms of sulfide so we use word \"penta\"
Solution
The name of As2S5 is diarsenic pentasulfide.
TRUE
Explanation
As there are 2 atoms of Arsenic so we uise word\"di\"
As there are 5 atoms of sulfide so we use word \"penta\".
The main thing we have to do is correct tap on the system status not.pdfappsmobileshoppe
The main thing we have to do is correct tap on the system status notice symbol and open the
Network and Sharing Center from the setting menu.
At that point tap on the Change connector settings hyperlink on the left-hand side of the Network
and Sharing Center.
Presently right-tap on the system connector you wish to change the DNS settings for and select
properties from the setting menu.
At the point when the properties for your system connector open, you should choose Internet
Protocol Version 4 (TCP/IPv4) from the rundown, then snap properties.
Our illustration will have opened the IPv4 settings for your system connector. You will find in
the base portion of the discourse that your PC is set to get its DNS settings \"consequently\". You
should change that with the goal that we can physically determine the DNS servers we need to
utilize. Presently basically pick an administration underneath and enter the DNS server
addresses.
Google DNS
Favored: 8.8.8.8
Interchange: 8.8.4.4
OpenDNS
Favored: 208.67.222.222
Interchange: 208.67.220.220
We selected to run with Google DNS for the time being — while they could utilize this
information to track your perusing propensities, by and by we\'re not that stressed over
Solution
The main thing we have to do is correct tap on the system status notice symbol and open the
Network and Sharing Center from the setting menu.
At that point tap on the Change connector settings hyperlink on the left-hand side of the Network
and Sharing Center.
Presently right-tap on the system connector you wish to change the DNS settings for and select
properties from the setting menu.
At the point when the properties for your system connector open, you should choose Internet
Protocol Version 4 (TCP/IPv4) from the rundown, then snap properties.
Our illustration will have opened the IPv4 settings for your system connector. You will find in
the base portion of the discourse that your PC is set to get its DNS settings \"consequently\". You
should change that with the goal that we can physically determine the DNS servers we need to
utilize. Presently basically pick an administration underneath and enter the DNS server
addresses.
Google DNS
Favored: 8.8.8.8
Interchange: 8.8.4.4
OpenDNS
Favored: 208.67.222.222
Interchange: 208.67.220.220
We selected to run with Google DNS for the time being — while they could utilize this
information to track your perusing propensities, by and by we\'re not that stressed over.
The answer is b. H2SBoth S and O come from the same group of the .pdfappsmobileshoppe
The answer is: b. H2S
Both S and O come from the same group of the periodic table and have 6 valence electrons.
Thus the structures of H2S and H2O are most similar to each other.
Solution
The answer is: b. H2S
Both S and O come from the same group of the periodic table and have 6 valence electrons.
Thus the structures of H2S and H2O are most similar to each other..
Molarity , M = ( massMolar mass) Volume of sol.pdfappsmobileshoppe
Molarity , M = ( mass/Molar mass) / Volume of solution in L = ( 5.00 g / 15.00
g/mol ) / 0.100 L = 3.3333 M
Solution
Molarity , M = ( mass/Molar mass) / Volume of solution in L = ( 5.00 g / 15.00
g/mol ) / 0.100 L = 3.3333 M.
Many molecules contain bonds that fall between th.pdfappsmobileshoppe
Many molecules contain bonds that fall between the extremes of ionic and covalent
bonds. The difference between the electronegativities of the atoms in these molecules is large
enough that the electrons aren\'t shared equally, and yet small enough that the electrons aren\'t
drawn exclusively to one of the atoms to form positive and negative ions. The bonds in these
molecules are said to be polar, because they have positive and negative ends, or poles, and the
molecules are often said to have a dipole moment. for Eg : HCl molecules, for example, have a
dipole moment because the hydrogen atom has a slight positive charge and the chlorine atom has
a slight negative charge. Because of the force of attraction between oppositely charged particles,
there is a small dipole-dipole force of attraction between adjacent HCl molecules. non-polar
molecule has it\'s electron density shifted by a charged molecule, i.e., if a positively charged
species is brought near a species with an evenly distributed charge, the electrons go towards the
side where the positive species is, and away from the opposite side, creating an artificial dipole.
Eg : a helium atom is perfectly symmetrical. But movement of the electrons around the nuclei of
a pair of neighboring helium atoms can become synchronized so that each atom simultaneously
obtains an induced dipole moment.
Solution
Many molecules contain bonds that fall between the extremes of ionic and covalent
bonds. The difference between the electronegativities of the atoms in these molecules is large
enough that the electrons aren\'t shared equally, and yet small enough that the electrons aren\'t
drawn exclusively to one of the atoms to form positive and negative ions. The bonds in these
molecules are said to be polar, because they have positive and negative ends, or poles, and the
molecules are often said to have a dipole moment. for Eg : HCl molecules, for example, have a
dipole moment because the hydrogen atom has a slight positive charge and the chlorine atom has
a slight negative charge. Because of the force of attraction between oppositely charged particles,
there is a small dipole-dipole force of attraction between adjacent HCl molecules. non-polar
molecule has it\'s electron density shifted by a charged molecule, i.e., if a positively charged
species is brought near a species with an evenly distributed charge, the electrons go towards the
side where the positive species is, and away from the opposite side, creating an artificial dipole.
Eg : a helium atom is perfectly symmetrical. But movement of the electrons around the nuclei of
a pair of neighboring helium atoms can become synchronized so that each atom simultaneously
obtains an induced dipole moment..
molality = moles of solute mass of solvent = 4..pdfappsmobileshoppe
molality = moles of solute / mass of solvent = 4.47 mol of CaCl2 / 1 g of water
mole fraction = moles of CaCl2 / (moles of CaCL2 + moles of water) = 4.47 / (4.47 + 1/18) =
0.987
Solution
molality = moles of solute / mass of solvent = 4.47 mol of CaCl2 / 1 g of water
mole fraction = moles of CaCl2 / (moles of CaCL2 + moles of water) = 4.47 / (4.47 + 1/18) =
0.987.
SolutionTrueIn a shared block prevents need another user from p.pdfappsmobileshoppe
Solution
:
True
In a shared block prevents need another user from performing DDL or DML operations on the
table.
The shared lock is automatically obtained when the select...for Update command is executed.
There is a lock known as share row exclusive, when this is lock is used then the whole table can
be seen by the other user. Similarly the other users can see all the rows in the table, but they
cannot perform DDL or DML operations on the table. When this is used in the shared mode then
it will automatically locks from updating of the table..
shares.txt(Save the file in the D drive.Then the path of the file po.pdfappsmobileshoppe
shares.txt(Save the file in the D drive.Then the path of the file pointing to that file is
D:\\\\shares.txt )
williams
kane
500
15
18
________________________________________________
C++ code
#include
#include
using namespace std;
int main()
{
//Declaring variables
int noOfShares;
double purPrice,sellPrice;
string firstname,lastname,fname,lname;
double
no_of_shares_sold,purchase_price,selling_price,tot_amt_invested,amount_received_after_sell,se
rvice_tax_during_buy,service_tax_during_sell,total_tax,amount_profit_lost;
//fileInput reference
ifstream dataIn;
//Opening input file
dataIn.open(\"D:\\\\shares.txt\");
//Reading the data from the file
while(dataIn>>fname>>lname>>noOfShares>>purPrice>>sellPrice)
{
firstname=fname;
lastname=lname;
no_of_shares_sold=noOfShares;
purchase_price=purPrice;
selling_price=sellPrice;
}
//closing the input stream
dataIn.close();
//calculating total amount invested
tot_amt_invested=no_of_shares_sold*purchase_price;
//Calculating service tax during buying of shares
service_tax_during_buy=tot_amt_invested*0.015;
//calculating the amount received after selling
amount_received_after_sell=no_of_shares_sold*selling_price;
//Calculating service tax after selling of shares
service_tax_during_sell=amount_received_after_sell*0.015;
//calculating Total Service tax paid
total_tax=service_tax_during_buy+service_tax_during_sell;
//calculating amount profit or lost
amount_profit_lost=(amount_received_after_sell-tot_amt_invested)-(total_tax);
//fileoutput reference
ofstream myfile;
//Opening the file
myfile.open (\"D:\\\\SharesOutput.txt\");
//Writing the contents to the file
myfile
<<\"No_Of_Shares_sold\"<<\"||\"<<\"Purchase_Price\"<<\"||\"<<\"Sell_Price\"<<\"||\"<<\"Amt_i
nvest\";
myfile<<\"||\"<<\"service_charges\"<<\"||\"<<\"Amt_Gained\"<<\"||\"<<\"Amt_received_After_s
ell\ \";
myfile<<\"\\t\"<
Solution
shares.txt(Save the file in the D drive.Then the path of the file pointing to that file is
D:\\\\shares.txt )
williams
kane
500
15
18
________________________________________________
C++ code
#include
#include
using namespace std;
int main()
{
//Declaring variables
int noOfShares;
double purPrice,sellPrice;
string firstname,lastname,fname,lname;
double
no_of_shares_sold,purchase_price,selling_price,tot_amt_invested,amount_received_after_sell,se
rvice_tax_during_buy,service_tax_during_sell,total_tax,amount_profit_lost;
//fileInput reference
ifstream dataIn;
//Opening input file
dataIn.open(\"D:\\\\shares.txt\");
//Reading the data from the file
while(dataIn>>fname>>lname>>noOfShares>>purPrice>>sellPrice)
{
firstname=fname;
lastname=lname;
no_of_shares_sold=noOfShares;
purchase_price=purPrice;
selling_price=sellPrice;
}
//closing the input stream
dataIn.close();
//calculating total amount invested
tot_amt_invested=no_of_shares_sold*purchase_price;
//Calculating service tax during buying of shares
service_tax_during_buy=tot_amt_invested*0.015;
//calculating the amount received after selling
amount_received_after_sell=no_of_shar.
Scientific approaches have been defined to depict the methods to stu.pdfappsmobileshoppe
Scientific approaches have been defined to depict the methods to study animal behaviour. These
methods help in understanding not only animal behaviour but also enhance the understanding of
an investigator in order to ddevelop a more comprehensive scienctific approach towards
psychological facts associated with animals. These methods are discussed as below:
1. The Ethology-Technically, ethology is the study of animal/human behaviuor in its
geographical location in association with its psycho-social environment with other factors by
means of close observations. This includes observational analysis by field scans i.e. residing in
the field/animal niche by the investigator for close observations. Importantly, ethology also
comprises of a biological context with each and every aspect of human/animal behaviuor. This
represents one of the most improtant tools to study and understand animal behaviour.
2. Comparative psychology- It is a comprehensive method which includes artificially designed
algorithm based non-human in silico/in vivo animals as models with special relevance for their
corresponding human behaviuor. This implies that a pre-defined set of rules are designed for a
model and certain stimuli are given. The nature of comprehensive behaviour of the organismic
model is then recorded and analysed. This gives a clear idea about comparative psychology of an
animal with respect to humans.
3. Behavioral studies and ecology- These studies comprise of a close integration of various
behavioral aspectso an animal with its specific environment. This field of science has been raised
in a past few decades in order to enhance the psychological information about the animal
behaviour. This helps the investigator to presume the behaviuor of an animals under a stimulus
without putting the animal exactly under the stimulus based upon previous experience,
recordings and integrated information.
Thus, collectively, these set of informations can help an investigator to understand animal
behaviuor scienctifically and psycho-socially.
Solution
Scientific approaches have been defined to depict the methods to study animal behaviour. These
methods help in understanding not only animal behaviour but also enhance the understanding of
an investigator in order to ddevelop a more comprehensive scienctific approach towards
psychological facts associated with animals. These methods are discussed as below:
1. The Ethology-Technically, ethology is the study of animal/human behaviuor in its
geographical location in association with its psycho-social environment with other factors by
means of close observations. This includes observational analysis by field scans i.e. residing in
the field/animal niche by the investigator for close observations. Importantly, ethology also
comprises of a biological context with each and every aspect of human/animal behaviuor. This
represents one of the most improtant tools to study and understand animal behaviour.
2. Comparative psychology- It i.
Ques-1Government funding is limited so that the factors are neede.pdfappsmobileshoppe
Ques-1:
Government funding is limited so that the factors are needed to consider to eliminate an
infectious disease are whether the infectious agent mode of transmission and infectious agent
etiology and disease symptoms in human beings. These factors are going to promote
\"epidemiologists\" to better control \"both nosocomial infections\", pandemic infections of
communicable disease. The possible preventive measures such as vaccination can be
implemented to eliminate the disease.
Assessment of chronic disease has not only been possible because of epidemiology but also acute
disease conditions, causes, outcome of treatment to that disease conditions is also possible by
studying the disease status in the public communities. This epidemiology study used to mainly
for those infectious diseases prevailed to a large number of populations within a geographical
area in a short period of time finally useful to implement public health policies against that
infection based on epidemiological data; for example infectious disease often prevail in a
population in 2 weeks. Example of epidemic diseases is meningococcal infections; Epidemiology
studies also useful to study the nature of pandemic diseases to monitor the health of communities
& populations; These studies are used for those diseases occurring in large number of
populations across the world in almost all geographical areas. Example is flu infections.
Ques-10:
No, we never win war on microbes because microbes can change their genome persistently &
actively through mutations when they reach the host genome. For example, HIV -1 infection
therefore, it is difficult to introduce medication to defeat this microbe
Some diseases causing bacterial spores are dormant for 1000 years also and medication is
ineffective to combat these spores due to the following reason
Bacillus subtilis produce dormant spores by sporulation ate suitable conditions and these spores
are called as endospores. These endopores have the capability to survive at extreme temperatures
followed by insufficient nutrient availability and their spore DNA is completely resistant to
mutagens such as UV light, chemicals. They are also resistant to desiccation and resistant to
extreme freezing and chemical disinfectants.
Factor-1: These endospores possess very complex diversified structure that can withstand all the
possible mutation -inducing changes. These possess structural adaptations such as exosporium,
hard spore coat and these are acts as very hard complex resistant materials, which can eliminate
digesting enzymes such as lysozymes. These endospores possess cortex composed of
peptidoglycan and it has a core wall that is present under the cortex and surrounds protoplast.
These layers potentially protect spore DNA material.
Factor-2: Core material of the spore also possesses calcium dipicolinate that enable stability of
chromosomal DNA by protecting it from oxidizing agents. Inside the core material, spore
possess DNA and it is covered .
Please follow the data and description Impact of absorption on tr.pdfappsmobileshoppe
Please follow the data and description :
Impact of absorption on transmission ranges :
In general the sound, light needs a medium to travel along the desination. Witht the transmission
there might be some level of difference or the deprecation at the destination and source
comparitively. This ranges from the level, range, medium and all other factors. These also vary
extensively based on the distances greater than about a hundred meters from the source. This
variation is caused by changes in weather conditions and by topographical features such as
ground cover, hills and other obstacles between the source and the receiver.
There are several important factors which affect the propagation such as geometric spreading,
atmospheric effects, and surface effects as discussed below :
a) Geometric Spreading :
This generally refers to the spreading of energy as a result of the expansion of the wavefronts.
Geometric spreading is merely independent of FREQUENCY and has a major effect in almost
all the propagation situations.
b) Air Absorption :
In general there are two types where the energy is absorbed by the atmosphere. These are
molecular relaxation and viscosity effects. High frequencies are absorbed more than low. The
amount of ABSORPTION depends on the temperature and humidity of the atmosphere.
c) Wind and Temperature Gradients :
The speed that the energy propagates in a gas depends on the temperature of the gas. Higher
temperatures produce higher speeds of sound. Since the temperature of the atmosphere is not
uniform there are local variations in the sound speed. For example, under normal conditions the
atmosphere is cooler at higher altitudes. This results in the waves being \'bent\' upwards. This
will result in the formation of a shadow zone, which is a region in which sound does not
penetrate. In reality some sound will enter this zone due to scattering. Scattering occurs when
sound waves are propagating through the atmosphere and meet a region of inhomogeneity and
some of their energy is re-directed into many other directions.
d) Ground Absorption :
This is to be noted in general as the energy is propagating over the ground, and then the
ATTENUATION will occur due to acoustic energy losses on REFLECTION. These losses will
depend on the surface. Smooth, hard surfaces will produce little ABSORPTION whereas thick
grass may result in sound levels being reduced by up to about 10 dB per 100 meters at 2000 Hz.
High frequencies are generally attenuated more than low frequencies.
These are some of the effects or the types where the absorption impacts the transmission of the
data or the energy from the source to the destination.
Hope this is helpful.
Solution
Please follow the data and description :
Impact of absorption on transmission ranges :
In general the sound, light needs a medium to travel along the desination. Witht the transmission
there might be some level of difference or the deprecation at the destination and source
comparitively. This rang.
Part-aLactose is a disaccharide and composed of glucose and galac.pdfappsmobileshoppe
Part-a:
Lactose is a disaccharide and composed of glucose and galactose through a glycosidic linkage
Part-b:
Process of lactose digestion & absorption:
Simple digestion of glucose rich carbohydrates by salivary amylase, pancreatic amylase and
intestinal \"maltase, sucrase and lactase\" into simple monomers of glucose units
Lactose digestion without lactose intolerance: It is common disaccharide that is present in the
milk and dairy products as it can be digested internally by lactase at temperature 37 °C and at pH
of 6. This is representing that enzymes specifically exerts their action at a specific temperature
and PH. Beta glycosidic bonds of lactose digested and forms D-galactose and D-glucose
individual units. Thereby substrate modification of the lactose by enzyme lactase revealing that
3’-OH and 2’-OH groups of the substrate galactopyranose component is very crucial for
competitive lactase enzymatic binding in which predominantly 3’-OH is essential for enzymatic
hydrolysis.
C12H22O11 (lactose) + H2O (enzymatic hydrolysis by lactase) C6H12O6 (D-galactose) +
C6H12O6 (D-glucose)+ heat.
In the absence of lactases in the small intestine is leading to lactose digestion by gut bacteria
through \"bacterial fermentation\" finally result in formation of lactic acid and CO2 result in
persistent flatulence and abdominal distension
Part-c:
Osmotic gradient is produced due to lactose intolerance in the large intestine through lactic acid
formation and bicarbonate formation between the \"lacteals of intestine\" and gut result in higher
fluid secretion into the intestine to attain homeostasis finally loose fecal formation or diarrhea.
Lactose intolerance: This is defined as a disorder in which inability of the production of enzyme
lactase to properly digest lactose in both adults and children as well, when they had dietary
intake of lactose containing food products (ex. milk). This condition is considered as congenital
since the gene coding for the enzyme lactose to break lactose sugar polymer
Solution
Part-a:
Lactose is a disaccharide and composed of glucose and galactose through a glycosidic linkage
Part-b:
Process of lactose digestion & absorption:
Simple digestion of glucose rich carbohydrates by salivary amylase, pancreatic amylase and
intestinal \"maltase, sucrase and lactase\" into simple monomers of glucose units
Lactose digestion without lactose intolerance: It is common disaccharide that is present in the
milk and dairy products as it can be digested internally by lactase at temperature 37 °C and at pH
of 6. This is representing that enzymes specifically exerts their action at a specific temperature
and PH. Beta glycosidic bonds of lactose digested and forms D-galactose and D-glucose
individual units. Thereby substrate modification of the lactose by enzyme lactase revealing that
3’-OH and 2’-OH groups of the substrate galactopyranose component is very crucial for
competitive lactase enzymatic binding in which predominantly 3’-OH is essential f.
In the first, the NO2 attaches at the 3 position .pdfappsmobileshoppe
In the first, the NO2 attaches at the 3 position In second, NO2 becomes NH2 In
third it becomes NN In fourth, it becomes the first diol again
Solution
In the first, the NO2 attaches at the 3 position In second, NO2 becomes NH2 In
third it becomes NN In fourth, it becomes the first diol again.
Labels on the right side of image (when we look at it)1. medial le.pdfappsmobileshoppe
Labels on the right side of image (when we look at it)
1. medial leminiscus
2.medulla oblongata
3. dorsal root ganglion
Labels on the left:
1. ventral nuclei
2. nucleus gracilis and nucleus cuneatus
3. fasciculus gracilis and fasciculus cuneatus
Solution
Labels on the right side of image (when we look at it)
1. medial leminiscus
2.medulla oblongata
3. dorsal root ganglion
Labels on the left:
1. ventral nuclei
2. nucleus gracilis and nucleus cuneatus
3. fasciculus gracilis and fasciculus cuneatus.
joint distribution function of X and Y will be given by int-( i.pdfappsmobileshoppe
joint distribution function of X and Y will be given by int->( infite to infinte ) int ->( infite to
infinte)g(x,y)dydx
int->ingration means
but here the values are between 0 to 1
int->( 0 to 1)int ->( 0 to 1)(Y-X)dXdY
int->( 0 to1) (0 to 1) (YX-X^2 / 2)= int ->( 0 to 1) (Y-1/2)dY=y^2/2-y=1/2-1=-1/2
Solution
joint distribution function of X and Y will be given by int->( infite to infinte ) int ->( infite to
infinte)g(x,y)dydx
int->ingration means
but here the values are between 0 to 1
int->( 0 to 1)int ->( 0 to 1)(Y-X)dXdY
int->( 0 to1) (0 to 1) (YX-X^2 / 2)= int ->( 0 to 1) (Y-1/2)dY=y^2/2-y=1/2-1=-1/2.
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
Welcome to TechSoup New Member Orientation and Q&A (May 2024).pdfTechSoup
In this webinar you will learn how your organization can access TechSoup's wide variety of product discount and donation programs. From hardware to software, we'll give you a tour of the tools available to help your nonprofit with productivity, collaboration, financial management, donor tracking, security, and more.
Unit 8 - Information and Communication Technology (Paper I).pdfThiyagu K
This slides describes the basic concepts of ICT, basics of Email, Emerging Technology and Digital Initiatives in Education. This presentations aligns with the UGC Paper I syllabus.
2024.06.01 Introducing a competency framework for languag learning materials ...Sandy Millin
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
Embracing GenAI - A Strategic ImperativePeter Windle
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
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.
Instructions for Submissions thorugh G- Classroom.pptxJheel Barad
This presentation provides a briefing on how to upload submissions and documents in Google Classroom. It was prepared as part of an orientation for new Sainik School in-service teacher trainees. As a training officer, my goal is to ensure that you are comfortable and proficient with this essential tool for managing assignments and fostering student engagement.
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.
Answer part a)According to the gene map of the wild type and the m.pdf
1. 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.
2. 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
