Heterozygous FH means a person has inhrited the fsulty gene from eit.pdf
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Heterozygous FH means a person has inhrited the fsulty gene from either of his/her parents. Homozygous FH-condition where a person inherits two faulty oraltered grnes from both the parents. If the two faulty genes are identical it is refereed to as homozygous FH. By calculating the chance by Punnett square- X+-faulty gene 2 children -Normal FH by inheriting one faulty and one healthy gene-XX+ and XY+ 1Child-Homozygous by receiving faulty gene from each of the parents-X+X+ 1Child-Normal-i.e. will not inherit FH as the child will receive the norml genes from the parents.XY Therefore, in one family of four chldren the ratio of homozygous FH will be 1:4, it will bw about 1 in a million.X+YX+X+X+X+YXXX+XY Solution Heterozygous FH means a person has inhrited the fsulty gene from either of his/her parents. Homozygous FH-condition where a person inherits two faulty oraltered grnes from both the parents. If the two faulty genes are identical it is refereed to as homozygous FH. By calculating the chance by Punnett square- X+-faulty gene 2 children -Normal FH by inheriting one faulty and one healthy gene-XX+ and XY+ 1Child-Homozygous by receiving faulty gene from each of the parents-X+X+ 1Child-Normal-i.e. will not inherit FH as the child will receive the norml genes from the parents.XY Therefore, in one family of four chldren the ratio of homozygous FH will be 1:4, it will bw about 1 in a million.X+YX+X+X+X+YXXX+XY.
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With X-linked inheritance, the X chromosome is transmitted in a diffe.pdf
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..
Answer ½Colorblindness is an X linked recessive inheritance. Affe.pdf
Answer: ½
Colorblindness is an X linked recessive inheritance. Affected father passes on the chromosome
to all of his daughters.
For a carrier female, with each pregnancy there is a one in two (50%) chance her sons will
inherit the disease allele and a one in two (50%) chance her daughters will be carriers.
GrandFather: XcY
XcY (father) x XcX (mother)
Son: XcY or XY
Hence: 50% probability or 1/2
Solution
Answer: ½
Colorblindness is an X linked recessive inheritance. Affected father passes on the chromosome
to all of his daughters.
For a carrier female, with each pregnancy there is a one in two (50%) chance her sons will
inherit the disease allele and a one in two (50%) chance her daughters will be carriers.
GrandFather: XcY
XcY (father) x XcX (mother)
Son: XcY or XY
Hence: 50% probability or 1/2.
X-Linked Recessive Write three rules to keep in mind when counseling .pdf
X-Linked Recessive Write three rules to keep in mind when counseling someone about the
likelihood of inheriting an X-linked recessive condition: About parents\' genotype. About
parents\' phenotype. About the probability of the offspring showing the trait. If instead of having
Hemophilia C (autosomal recessive), Lisa Keller\'s brother had Hemophilia A (X-linked
recessive), what would be the probability that she was a carrier? 25% 50% 67% 100% More
information is required to answer this question.
Solution
About genotype of parents- If the mother is a carrier i.e. carrying the recessive allele on one of
her X chromosome and the father do not have the recessive allele in his genotype, then none of
the daughters will suffer from the disease (since they will inherit only one recessive allele from
the mother) but sons will have a 50% probability of suffering from the disease as they will
inherit the recessive allele from the mother and the Y chromosome from the father, although
there is a 50% probability of inheriting the wild type allele from the mother.
Now if the mother is suffering from the disease i.e. has the recessive allele on both the X
chromosome while the father is healthy having no allele for the disease, the daughters will
always be a carrier but never suffer from the disease as they eill inherit the healthy wild type
allele from the father as one X chromosome comes from the father and the other X chromosome
from the mother. The sons will however all suffer from the disease as they inherit the Y
chromosome from the father and the X chromosome having the recessive allele causing the
disease will be inherited from the mother.
Now suppose the mother is a carrier i.e. carries the recessive allele only on one X chromosome
but the father suffers from the disease making his genotype XrY (Xr denoting recessive allele
carrying X chromosome), then, 50%of their son will duffer from the disease while 50% of the
daughters will also have a chance of suffering from the disease.
If both the parents do not have a single recessive X linked disease gene on any of their X
chromosome then none of the children will suffer from the disease.
About phenotype of parents- If the mother is a carrier i.e. have the diseased allele on only one X
Chromosome, then the phenotype of mother will be normal. But the father can never be a carrier
for a X linked recessive disease as males have only one X chromosome. If the diseased allele in
present on that X chromosome in males it will be expressed as there is no dominant healthy wild
type allele to dominate the diseased gene. Thus if the father has the recessive diseased allele on
X chromosome, the father\'s phenotype will be expresding the disease while if the father do not
have the allele he will be healthy. On the other hand for the mother to express the disease the
recessive allele has to be present on bith her X chromosome.
Probability of offsprings showing the trait- If mother is a carrier and father is normal, then 5.
Consider the following pedigree, which traces the inheritance of a si.pdf
Consider the following pedigree, which traces the inheritance of a single-gene hereditary
disease. Only individuals that are either affected or normal are shown. In other words, potential
heterozygotes are NOT indicated. Characterize each of the following modes of inheritance as:
impossible, unlikely, or probable. Justify your answers. Autosomal dominant Autosomal
recessive X-linked dominant X-Iinked recessive Y linked
Solution
Single gene herediary diseases are mainly caused by a change of any one gene in a particular
DNA.Its divided into diiferemt categories like dominant ,recessive and X linked.Square indicates
a male and circle female.According to the diagram affected male is crossed with an unaffected
female.The progeny obtained is affected daughters and unaffected sons.
Autosomal dominant: in this case, the father — has a 50 percent chance of having an affected
child with one mutated gene (dominant gene) and a 50 percent chance of having an unaffected
child with two normal genes (recessive genes).But, in our case none of the sons are affected
which means its criss cross inheritance that is from father to daughter.Hence , this case is
unlikely
Autosomal recessive:To have an autosomal recessive disorder, you inherit two mutated genes,
one from each parent. These disorders are usually passed on by two carriers.The cross given in
the question is not between the two carriers.The male parent is affected for sure hence this case is
impossible
X linked dominant:As in autosomal dominant inheritance, only one copy of a disease allele on
the X chromosome is required for an individual to be susceptible to an X-linked dominant
disease.Both males and females can be affected, although males may be more severely affected
because they only carry one copy of genes found on the X chromosome. Some X-linked
dominant disorders are lethal in males.When a female is affected, each pregnancy will have a
one in two (50%) chance for the offspring to inherit the disease allele. When a male is affected,
all his daughters will be affected, but none of his sons will be affected.So, there are probable
chance for this type of inheritance
X linked recessive: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. Each daughter will have a 50% chance of being an unaffected
carrier and a 50% chance of both X chromosomes being normal.In our cross all daughters are
affected which means this type of inheritance is impossible
Y linked:This type of inheritance is impossible in this case as all female children are getting
affected.If it would have been Y linked then only sons should have been affected as Y
chromosome is present only in males.Hence this.
If the F1 plants with red flowers were crossed to each other, what pr.pdf
If the F1 plants with red flowers were crossed to each other, what progeny would be produced
and in what proportions? On occasion, genetically identical humans are produced by a natural
cloning process. The result is identical, or monozygotic, twins. These individuals have all 23
pairs of chromosomes identical to those of their twin. Fraternal twins are produced when two
eggs are ovulated and fertilized simultaneously, but independently. What is the likelihood that
two fraternal twins will be genetically identical? On average, what proportion of their alleles do
siblings have in common with each other? On average, what is the probability that any particular
X-linked allele carried by a man is also carried by his brother? On average, what proportion of
their alleles do children have in common with a parent? You work as a genetics consultant for a
local attorney. She is currently working on five cases of disputed paternity. Tests have produced
the following information on blood types:
Solution
2A. Fraternal twins are formed from the fertilization of two different ovules by two different
sperms and are hence also known as \'dizygotic\'. This implies that there is hardly any likelihood
of being any more identical than two siblings, which is 50 percent. However, they can be formed
either due to the presence of hyperovulation gene in the mother, which makes the mother release
more than one egg at a time, or by multiple fathers within a span of few days. In the first case,
the likehood of being identical is 50 percent or more, whereas in the second case, it\'s only the
mother\'s gene which is similar in the twins.
2B. Each sibling gets half an allele from each parent, so 50% is all they get from the parents. So
one extreme can be each sibling getting the alleles from parents and which the other didn\'t get,
while the other extreme can be each sibling getting exactly the same allele. Hence the probability
ranges from 0 (no similarity) to 1 (exactly same). So on a average, the proportion of alleles they
share is (0+1)/2 = 0.5.
2C. For a man and his brother, the X linked genes can only come from their mother since they
will recieve Y gene from their father. Hence, the likelihood is 50%. However, to get the same X-
linked genes from mother, the likelihood reduces to 25% (Please refer to the table below).
X1-M, X-F (25%)
(Daughter)
X2-M, X-F (25%)
(Daughter)
X1-M, Y-F (25%)
(Son)
X2-M, Y-F (25%)
(Son)
2D. Since each child receives only half allele from each parent, on an average the have only 50%
common with a parent.MotherMotherX1-MX2-MFatherX-F
X1-M, X-F (25%)
(Daughter)
X2-M, X-F (25%)
(Daughter)FatherY-F
X1-M, Y-F (25%)
(Son)
X2-M, Y-F (25%)
(Son).
Heritable Traits in Man, Pedigree Analysis and Pedigree Application
This presentation is required in our Genetics Subject
Recommended
With X-linked inheritance, the X chromosome is transmitted in a diffe.pdf
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..
Answer ½Colorblindness is an X linked recessive inheritance. Affe.pdf
Answer: ½
Colorblindness is an X linked recessive inheritance. Affected father passes on the chromosome
to all of his daughters.
For a carrier female, with each pregnancy there is a one in two (50%) chance her sons will
inherit the disease allele and a one in two (50%) chance her daughters will be carriers.
GrandFather: XcY
XcY (father) x XcX (mother)
Son: XcY or XY
Hence: 50% probability or 1/2
Solution
Answer: ½
Colorblindness is an X linked recessive inheritance. Affected father passes on the chromosome
to all of his daughters.
For a carrier female, with each pregnancy there is a one in two (50%) chance her sons will
inherit the disease allele and a one in two (50%) chance her daughters will be carriers.
GrandFather: XcY
XcY (father) x XcX (mother)
Son: XcY or XY
Hence: 50% probability or 1/2.
X-Linked Recessive Write three rules to keep in mind when counseling .pdf
X-Linked Recessive Write three rules to keep in mind when counseling someone about the
likelihood of inheriting an X-linked recessive condition: About parents\' genotype. About
parents\' phenotype. About the probability of the offspring showing the trait. If instead of having
Hemophilia C (autosomal recessive), Lisa Keller\'s brother had Hemophilia A (X-linked
recessive), what would be the probability that she was a carrier? 25% 50% 67% 100% More
information is required to answer this question.
Solution
About genotype of parents- If the mother is a carrier i.e. carrying the recessive allele on one of
her X chromosome and the father do not have the recessive allele in his genotype, then none of
the daughters will suffer from the disease (since they will inherit only one recessive allele from
the mother) but sons will have a 50% probability of suffering from the disease as they will
inherit the recessive allele from the mother and the Y chromosome from the father, although
there is a 50% probability of inheriting the wild type allele from the mother.
Now if the mother is suffering from the disease i.e. has the recessive allele on both the X
chromosome while the father is healthy having no allele for the disease, the daughters will
always be a carrier but never suffer from the disease as they eill inherit the healthy wild type
allele from the father as one X chromosome comes from the father and the other X chromosome
from the mother. The sons will however all suffer from the disease as they inherit the Y
chromosome from the father and the X chromosome having the recessive allele causing the
disease will be inherited from the mother.
Now suppose the mother is a carrier i.e. carries the recessive allele only on one X chromosome
but the father suffers from the disease making his genotype XrY (Xr denoting recessive allele
carrying X chromosome), then, 50%of their son will duffer from the disease while 50% of the
daughters will also have a chance of suffering from the disease.
If both the parents do not have a single recessive X linked disease gene on any of their X
chromosome then none of the children will suffer from the disease.
About phenotype of parents- If the mother is a carrier i.e. have the diseased allele on only one X
Chromosome, then the phenotype of mother will be normal. But the father can never be a carrier
for a X linked recessive disease as males have only one X chromosome. If the diseased allele in
present on that X chromosome in males it will be expressed as there is no dominant healthy wild
type allele to dominate the diseased gene. Thus if the father has the recessive diseased allele on
X chromosome, the father\'s phenotype will be expresding the disease while if the father do not
have the allele he will be healthy. On the other hand for the mother to express the disease the
recessive allele has to be present on bith her X chromosome.
Probability of offsprings showing the trait- If mother is a carrier and father is normal, then 5.
Consider the following pedigree, which traces the inheritance of a si.pdf
Consider the following pedigree, which traces the inheritance of a single-gene hereditary
disease. Only individuals that are either affected or normal are shown. In other words, potential
heterozygotes are NOT indicated. Characterize each of the following modes of inheritance as:
impossible, unlikely, or probable. Justify your answers. Autosomal dominant Autosomal
recessive X-linked dominant X-Iinked recessive Y linked
Solution
Single gene herediary diseases are mainly caused by a change of any one gene in a particular
DNA.Its divided into diiferemt categories like dominant ,recessive and X linked.Square indicates
a male and circle female.According to the diagram affected male is crossed with an unaffected
female.The progeny obtained is affected daughters and unaffected sons.
Autosomal dominant: in this case, the father — has a 50 percent chance of having an affected
child with one mutated gene (dominant gene) and a 50 percent chance of having an unaffected
child with two normal genes (recessive genes).But, in our case none of the sons are affected
which means its criss cross inheritance that is from father to daughter.Hence , this case is
unlikely
Autosomal recessive:To have an autosomal recessive disorder, you inherit two mutated genes,
one from each parent. These disorders are usually passed on by two carriers.The cross given in
the question is not between the two carriers.The male parent is affected for sure hence this case is
impossible
X linked dominant:As in autosomal dominant inheritance, only one copy of a disease allele on
the X chromosome is required for an individual to be susceptible to an X-linked dominant
disease.Both males and females can be affected, although males may be more severely affected
because they only carry one copy of genes found on the X chromosome. Some X-linked
dominant disorders are lethal in males.When a female is affected, each pregnancy will have a
one in two (50%) chance for the offspring to inherit the disease allele. When a male is affected,
all his daughters will be affected, but none of his sons will be affected.So, there are probable
chance for this type of inheritance
X linked recessive: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. Each daughter will have a 50% chance of being an unaffected
carrier and a 50% chance of both X chromosomes being normal.In our cross all daughters are
affected which means this type of inheritance is impossible
Y linked:This type of inheritance is impossible in this case as all female children are getting
affected.If it would have been Y linked then only sons should have been affected as Y
chromosome is present only in males.Hence this.
If the F1 plants with red flowers were crossed to each other, what pr.pdf
If the F1 plants with red flowers were crossed to each other, what progeny would be produced
and in what proportions? On occasion, genetically identical humans are produced by a natural
cloning process. The result is identical, or monozygotic, twins. These individuals have all 23
pairs of chromosomes identical to those of their twin. Fraternal twins are produced when two
eggs are ovulated and fertilized simultaneously, but independently. What is the likelihood that
two fraternal twins will be genetically identical? On average, what proportion of their alleles do
siblings have in common with each other? On average, what is the probability that any particular
X-linked allele carried by a man is also carried by his brother? On average, what proportion of
their alleles do children have in common with a parent? You work as a genetics consultant for a
local attorney. She is currently working on five cases of disputed paternity. Tests have produced
the following information on blood types:
Solution
2A. Fraternal twins are formed from the fertilization of two different ovules by two different
sperms and are hence also known as \'dizygotic\'. This implies that there is hardly any likelihood
of being any more identical than two siblings, which is 50 percent. However, they can be formed
either due to the presence of hyperovulation gene in the mother, which makes the mother release
more than one egg at a time, or by multiple fathers within a span of few days. In the first case,
the likehood of being identical is 50 percent or more, whereas in the second case, it\'s only the
mother\'s gene which is similar in the twins.
2B. Each sibling gets half an allele from each parent, so 50% is all they get from the parents. So
one extreme can be each sibling getting the alleles from parents and which the other didn\'t get,
while the other extreme can be each sibling getting exactly the same allele. Hence the probability
ranges from 0 (no similarity) to 1 (exactly same). So on a average, the proportion of alleles they
share is (0+1)/2 = 0.5.
2C. For a man and his brother, the X linked genes can only come from their mother since they
will recieve Y gene from their father. Hence, the likelihood is 50%. However, to get the same X-
linked genes from mother, the likelihood reduces to 25% (Please refer to the table below).
X1-M, X-F (25%)
(Daughter)
X2-M, X-F (25%)
(Daughter)
X1-M, Y-F (25%)
(Son)
X2-M, Y-F (25%)
(Son)
2D. Since each child receives only half allele from each parent, on an average the have only 50%
common with a parent.MotherMotherX1-MX2-MFatherX-F
X1-M, X-F (25%)
(Daughter)
X2-M, X-F (25%)
(Daughter)FatherY-F
X1-M, Y-F (25%)
(Son)
X2-M, Y-F (25%)
(Son).
Heritable Traits in Man, Pedigree Analysis and Pedigree Application
This presentation is required in our Genetics Subject
If the characteristic followed in the pedigree is x-linked recessive .pdf
If the characteristic followed in the pedigree is x-linked recessive allele, what is III-1\'s
genotype? a. Hemizygous for a dominant allele b. Hemizygous for a recessive allele c.
Definitely heterozygous d. dominant e. Either heterozygous or homozygous dominant
Solution
1st generation parents 1,2 though are normal but must be heterozygotes (Pp) since their children
are affected i.e. 2nd generation males 1,3.
Since 1st generation male 3 is affected (pp) and 4 is normal but regardless if 4 in 1st generation
is PP or Pp their children will be either 4 Pp i.e. heterozygote or 2 affected pp and 2
heterozygotes Pp. But the 2nd generation 4, 5 female are normal but heterozygotes Pp.
So the cross between a 2nd generation affected male 3(pp) and a normal but heterozygote female
4(Pp) would have a normal male child only if hemizygous for a dominant allele condition exists.
Hemizygous is a condition in which only one gene copy/DNA sequence exists in diploid cells.
Males are hemizygous for most genes on sex chromosomes since they have only one X and one
Y chromosome unlike two X for females. A recessive X-linked gene requires two mutant alleles
to affect the female phenotype.
Rules followed here are.
Hemophilia is a recessive trait. The dominant trait is normal blood d.pdf
Hemophilia is a recessive trait. The dominant trait is normal blood dotting A gene that can cause
hemophilia is located on the X chromosome. If a homozygous normal female marries a normal
male, they children might have which of the following phenotypes: Select one: a. normal
females, hemophiliac females, and hemophiliac males. b. normal females, hemophiliac females,
normal males, and hemophiliac males. c. normal females and hemophiliac males d. normal
females and normal mates. e. normal females, normal males or hemophiliac males.
Solution
In X - linked recessive inheritance males are more prone to disease because they carry the
disease trait. If the affected offspring’s mother has brother or father containing the same trait
then the mother pass the trait to her offspring. The affected males produce unaffected daughters
who can produce 50% affected sons. Hemophilia is the example of X - linked recessive trait.
When a homozygous normal female marries to a normal male the offspring produced are normal
females and normal males. Option d is correct..
Ferdinand is affected by hemophilia, a recessive x-linked condition..pdf
Ferdinand is affected by hemophilia, a recessive x-linked condition. He mates with an unaffected
woman with no history of hemophilia in her family. Their first child is a girl with Turner
Syndrome (XO sex chromosomes), who is also hemophiliac. You infer that she resulted from the
following sperm and egg (answers listed in the order of sperm, egg):normal, normal
Solution
Answer: Because Ferdinand has hemophilia which is X-linked ressesive disease. Daughters
produced by mating of ferdinand and a normal female would be carriers of hemophilia. In this
case however, the daughter is having Turners\' syndrome (XO) and is heophilic as well. This
implies that the X came from sperm of Ferdinand and the O is beacuse the X chromosome from
the female was nondisjunct at meiosis I, as non-disjunction at meiosis II would result in two
gaetes being normal, One being trisomic and one being monosomic. whereas nondysjunction at
meiosis I would give rise to 50% trisomic and 50% monosomic gametes. Hence raising the
possibility of turners\' syndrome after nondysjunction at meiosis I.
Therefore answer is:
c. Normal and non-dysjunction at meiosis I..
Sex-linked inheritance by Puzon and Tope
Sex-linked inheritance by Puzon, kathleen Faye R. and Tope, Donita J.
Inbreeding and outbreeding
This powerpoint gives a clear picture on inbreeding and also about outbreeding of higher organisms. This also explains the advantages and disadvantages of the above said topics. the methods of inbreeding and reasons for inbreeding also given in this powerpoint.
1. In a given plant, the yellow seed color allele (Y) is dominant to .pdf
1. In a given plant, the yellow seed color allele (Y) is dominant to green seed color (y). Give the
genotypic and phenotypic ratios for the following crosses? a. homozygous yellow x green b.
heterozygous yellow x green c. heterozygous yellow x heterozygous yellow d. green x green 2. If
two animals heterozygous for a single pair of genes are mated and have 276 offspring, how many
would be expected to have the dominant genotype? 3. Two long-winged fies are mated. The
offspring included 90 with long wings and 32 with short wings. Which phenotype is recessive?
4. A blood test reveals that a couple both have the sickle-cell trat. What is the probability of their
producing a child with sickle-cell anemia? 5. A blood type A woman has a type B child. She
accuses a type O man of being the father. Knowing genetics you are asked to give an opinion; do
so. 6. Assume that human eye color is controlled by a single pair of alleles. A blue-eyed man,
both of whose parents were bom brown-eyed, marries a brown-eyed woman whose father was
blue-eyed and whose mother was brown-eyed. They have a blue-eyed child. Which are the
genotypes of all individuals mentioned? 7. A red fruit, tall tomato plant RrTt, is crossed with a
yellow-fruited dwarf plant ritt. What genotypes and phenotypes would be expected in the
offispring? 8. What genotype and phenotype would be expected in the offspring if the cross in
the previous problem had been RrTt x RrTt ?.
sex linked inheritance, Sex Influence inheritance and sex limited characters
sex linked inheritance, Sex Influence inheritance and sex limited characters
Answer Every investor expects dividend from his investments.Dividen.pdf
Answer: Every investor expects dividend from his investments.Dividend income helps investors
in dealing with their routine expenses. Investors put their money to earn some positive returns
and dividend is one of the way of getting returns for investors. So, It is a duty of management to
satisfy dividend expectation of investors.
If management do not pay dividend then it is known as retention of earnings. Retention earnings
has it\'s own pro and cons, which are described as follows:
Pros:
Retained earnings reduces organization\'s dependence on external borrowings. It can be
considered as an advantage when external borrowings are not available easily.
Retained earnings indicates good business opportunities, It means organization is growing
financially. It may result in better stock valuation of organization.
In some situation dividend is subject to taxes (either distribution tax or income tax) in the hands
of investors. So, in this case investors are not interested in dividend but they are more interested
in capital gain(which may be tax exempt). So, In this situation retained earnings fulfill investor\'s
objective.
Cons:
As explained earlier all the investors have some expectation of dividend, in case of higher
retention of earnings it may become difficult to fulfill investor\'s expectations.
Retention of earnings may affect investor\'s earning capacity when investors have available
opportunities, which gives higher return than organization earns by retaining amount. In this case
investor looses earning opportunities.
In case of dividend management\'s decision is significantly influenced by media and institutional
investors. Media is responsible for organization\'s image in society, such image significantly
affects organization\'s financial performance. This is why management always considers media\'s
point of view before making any dividend decision.
Institutional investors are usually invests their money and takes part in decision making process
of organization. Institutional investor\'s weight in decision making process is generally very
high, so it is obvious that they influence management\'s dividend decision also.
Solution
Answer: Every investor expects dividend from his investments.Dividend income helps investors
in dealing with their routine expenses. Investors put their money to earn some positive returns
and dividend is one of the way of getting returns for investors. So, It is a duty of management to
satisfy dividend expectation of investors.
If management do not pay dividend then it is known as retention of earnings. Retention earnings
has it\'s own pro and cons, which are described as follows:
Pros:
Retained earnings reduces organization\'s dependence on external borrowings. It can be
considered as an advantage when external borrowings are not available easily.
Retained earnings indicates good business opportunities, It means organization is growing
financially. It may result in better stock valuation of organization.
In some situ.
Ans. Gene is defined as the segment of DNA that gives a functional p.pdf
Ans. Gene is defined as the segment of DNA that gives a functional product (protein, RNA). A
functional product can be either protein (or polypeptide) or RNA.
mRNA transcribes from a gene is further translated to produce polypeptides. However, all other
RNAs are also transcribed from their respective genes. It includes tRNA, rRNA, snRNA and
various regulatory RNAs (micro RNA, small interfering RNA), etc. tRNA acts as amino-acid
carrier during translation. rRNA is the structural and functional scaffold of ribosomes. snRNA
mediates splicing of immature mRNA. miRNA and siRNA have crucial roles in gene
regulations.
SO, polypeptides are just one class of functional product from DNA. The genomic DNA also
produces various classes of RNAs with diverse functions that are crucial for cellular activates
and gene regulation.
Therefore, the statement “Add DNA sequences in the genome that does not encode polypeptides
can be defined as junk DNA” is INCORRECT. Junk DNA is non-coding DNA that excludes all
genes (including those encoding polypeptides and RNAs).
Solution
Ans. Gene is defined as the segment of DNA that gives a functional product (protein, RNA). A
functional product can be either protein (or polypeptide) or RNA.
mRNA transcribes from a gene is further translated to produce polypeptides. However, all other
RNAs are also transcribed from their respective genes. It includes tRNA, rRNA, snRNA and
various regulatory RNAs (micro RNA, small interfering RNA), etc. tRNA acts as amino-acid
carrier during translation. rRNA is the structural and functional scaffold of ribosomes. snRNA
mediates splicing of immature mRNA. miRNA and siRNA have crucial roles in gene
regulations.
SO, polypeptides are just one class of functional product from DNA. The genomic DNA also
produces various classes of RNAs with diverse functions that are crucial for cellular activates
and gene regulation.
Therefore, the statement “Add DNA sequences in the genome that does not encode polypeptides
can be defined as junk DNA” is INCORRECT. Junk DNA is non-coding DNA that excludes all
genes (including those encoding polypeptides and RNAs)..
additional optmization techniques for underlying IP network must1.pdf
additional optmization techniques for underlying IP network must:
1.Have bandwidth and performance to handle converged services.
2.Seprate VLAN for Voice traffic.
3.Meet the demand of high availability of voice services by providing resiliency to mitigate the
effect of network outages.
4.Should be modular, hierarchical & consistent to promote consistency and manageability.
5.configure QoS on core switch.
Solution
additional optmization techniques for underlying IP network must:
1.Have bandwidth and performance to handle converged services.
2.Seprate VLAN for Voice traffic.
3.Meet the demand of high availability of voice services by providing resiliency to mitigate the
effect of network outages.
4.Should be modular, hierarchical & consistent to promote consistency and manageability.
5.configure QoS on core switch..
According to the given equation, aqueous carbon dioxide reacts with .pdf
According to the given equation, aqueous carbon dioxide reacts with water to form carbonic
acid, Carbonic acid is not stable in normal body temperatures, therefore, it ionizes into
bicarbonate ion (HCO3-) and a hydrogen ion (H+) is liberated. The liberation of hydrogen ions
decreases the pH.
During breathing, oxygen is inhaled which is carried by the blood to the different tissues of the
body. Oxygen facilitates the various metabolic reactions occurring inside our cells and helps in
respiration, or the generation of energy. Carbon dioxide, formed as a by-product of these energy-
generating metabolic reactions, diffuses out of the tissues and into the blood stream, which
carries CO2 to the lungs, where it is exhaled out.
a) In case of hypoventilation :
The exchange of gases that occurs in the lungs while breathing is impaired. Therefore, the
requisite amount of carbon dioxide is not being exhaled out. This leads to a build-up of carbon-
dioxide in the blood plasma. As a result, the forward reaction is facilitated and more H+ ions are
generated. The higher concentration of H+ ions decreases the pH (acidic pH). This can lead to
acidosis.
The build-up of carbon dioxide in the blood facilitates the forward reaction and therefore the
equilibrium shifts to the right.
The hydronium ion or H3O+ is the product of a hydrogen ion (H+) and water. Since a hydrogen
ion is too small to exist freely in water, it therefore combines with a water molecule to form a
hydronium ion which is more stable. In case of hypoventilation, the concentration of H+ ions
increase. Therefore, the concentration of hydronium ions increase during hypoventilation.
The stress concentration or partial pressure of CO2 during hypoventilation is significantly higher
than 40mmHg (around 60-65 mmHg).
b) In case of hyperventilation :
The breathing rate is more than normal. Therefore, as a result of heavier and faster breathing, the
CO2 concentration in the blood rapidly decreases, that is, CO2 easily leaves the blood plasma.
Decrease in the CO2 concentration slows down the rate of the forward reaction, reducing the
levels at which bicarbonate ion and H+ ion concentration were being generated. The bicarbonate
and hydrogen ions recombine to yield CO2 and water. This lowering in hydrogen ion
concentration consequently lowers the pH (alkaline pH). This can lead to alkalosis.
As a result of the lowering of CO2 , the reverse reaction takes place. The carbonic acid changes
back to CO2 and H2O, shifting the equilibrium to the left.
The concentration of H3O+ decreases during hyperventilation, since the H+ ion concentration is
lowered.
The stress levels or partial pressure of CO2 during hyperventilation is significantly lower than 40
mmHg (around 25-30 mmHg).
c) If sodium carbonate or Na2CO3 is added instead of sodium bicarbonate, then sodium
carbonate would react with water and it would hydrolyse into Na+ and CO32- ions. The CO32-
combine with the hydrogen (H) present in water to form H2CO3 (carbon.
a) mean = 1.43Thus distribution is Poisson(4.2)P(X = 4) = 4.2^4.pdf
a) mean = 1.4*3
Thus distribution is Poisson(4.2)
P(X = 4) = 4.2^4 * e^{-4.2} / 4! = 0.1944
b)
average number of ghosts in 90min (1.5hrs) interval = 1.4*1.5 = 2.1
c)
Probability that 4th ghost will appear before 10 am
= Probability that number of ghosts appearing from 8 to 10 am >= 4
As it is a 2 hr period, the distribution of number of ghosts is Poisson(1.4*2) i.e Poisson(2.8)
P(X >= 4)
= 1 - P(X < 4)
= 1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3)]
= 1 - [(2.8^0 * e^{-2.8} / 0!) + (2.8^1 * e^{-2.8} / 1!) + (2.8^2 * e^{-2.8} / 2!) + (2.8^3 * e^{-
2.8} / 3!)]
= 1 - [0.0608 + 0.1703 + 0.2384 + 0.2225]
= 0.692
d)
When arrival distribution is Poisson(lambda), inter-arrival time distribution is
Exponential(lambda)
Thus, interarrival time distribution is Exponential(1.4)
Expected value of time between 5th and 6th ghost = 1/1.4 = 0.7143hr
e)
On average time taken for 10th ghost to arrive = 10*(1/1.4) = 7.143hr
Thus on an average, 10th ghost will arrive at 3:09pm
f)
Expected value of time between ghost appearances = 1/1.4 = 0.7143hr (see part d)
g)
Probability that 3rd ghost appears after 9:45 am
= Probablity that number of ghosts appearing till 9:45 am < 3
Duration from 8am to 9:45 am = 1.75hr
Mean number of ghosts = 1.4*1.75 = 2.45
Thus X is Poisson(2.45)
P(X < 3)
= P(X = 0) + P(X = 1) + P(X = 2)
= (2.45^0 * e^{-2.45} / 0!) + (2.45^1 * e^{-2.45} / 1!) + (2.45^2 * e^{-2.45} / 2!)
= 0.0863 + 0.2114 + 0.2590
= 0.5567
h)
As inter-arrival time is Exponential(1.4), inter-arrival time is memoryless
Thus, time for 7th ghost will be calculated from 1:00 pm and not from 12:35pm due to
memorylessness property.
Time duration from 1:00pm to 1:15pm = 0.25hrs
Let Y be random variable denoting inter-arrival time
probability that the 7th ghost will appear before 1:15 p.m.
= P(Y < 0.25)
= 1 - e^{-1.4 * 0.25)
= 0.2953
i)
Expected value of time for 7th ghost to appear after 1:00pm = 1/1.4 = 0.7143hr = 42.86 min
Time from 12:35pm to 1:00pm = 25min
Thus expected interarrival time = 25 + 42.86 = 67.86 min = 1hr 7.86min
j)
Expected duration after 1:00pm for 9th ghost to appear = 3 * (1/1.4) = 2.143hr = 2hr 8.58min
Thus expected time at which 9th ghost appears = 3:09pm
k)
As the inter-arrival time follows memorylessness property, the future probabilities do not change
with the observed value at past.
Thus, 4 ghosts appearing between 7:00pm to 10:00pm does not affect ghost appearance from
11:00pm to 11:30pm
Duration from 11:00pm to 11:30pm = 0.5hrs
mean number of ghosts = 1.4*0.5 = 0.7
Thus it follows Poisson(0.7) distribution
P(X > 2)
= 1 - P(X <= 2)
= 1 - [P(X = 0) + P(X = 1) + P(X = 2)]
= 1 - [(0.7^0 * e^{-0.7} / 0!) + (0.7^1 * e^{-0.7} / 1!) + (0.7^2 * e^{-0.7} / 2!)]
= 1 - [0.4966 + 0.3476 + 0.1217]
= 0.0341
l)
Probability that 8 ghosts appear between 6pm and midnight
= Probability that number of ghosts appear from 6pm to 7pm and 10pm to midnight = 4 as it is
known that 4 ghosts appeared from 7pm to 10pm
Let A denote number of ghost from 6pm to 7pm
B denote number of ghosts .
2.a. Wired Media Type and ExplinationTwisted-Pair CableTwiste.pdf
2.
a. Wired Media Type and Explination:
Twisted-Pair Cable
Twisted-pair cable is a type of cabling that is used for telephone communications and most
modern Ethernet networks. A pair of wires forms a circuit that can transmit data. The pairs are
twisted to provide protection against crosstalk, the noise generated by adjacent pairs. When
electrical current flows through a wire, it creates a small, circular magnetic field around the wire.
When two wires in an electrical circuit are placed close together, their magnetic fields are the
exact opposite of each other. Thus, the two magnetic fields cancel each other out. They also
cancel out any outside magnetic fields. Twisting the wires can enhance this cancellation effect.
Using cancellation together with twisting the wires, cable designers can effectively provide self-
shielding for wire pairs within the network media.
Two basic types of twisted-pair cable exist: unshielded twisted pair (UTP) and shielded twisted
pair (STP). The following sections discuss UTP and STP cable in more detail.
UTP Cable
UTP cable is a medium that is composed of pairs of wires . UTP cable is used in a variety of
networks. Each of the eight individual copper wires in UTP cable \\is covered by an insulating
material. In addition, the wires in each pair are twisted around each other.
UTP cable relies solely on the cancellation effect produced by the twisted wire pairs to limit
signal degradation caused by electromagnetic interference (EMI) and radio frequency
interference (RFI). To further reduce crosstalk between the pairs in UTP cable, the number of
twists in the wire pairs varies. UTP cable must follow precise specifications governing how
many twists or braids are permitted per meter (3.28 feet) of cable.
UTP cable often is installed using a Registered Jack 45 (RJ-45) connector The RJ-45 is an eight-
wire connector used commonly to connect computers onto a local-area network (LAN),
especially Ethernets.
When used as a networking medium, UTP cable has four pairs of either 22- or 24-gauge copper
wire. UTP used as a networking medium has an impedance of 100 ohms; this differentiates it
from other types of twisted-pair wiring such as that used for telephone wiring, which has
impedance of 600 ohms.
UTP cable offers many advantages. Because UTP has an external diameter of approximately
0.43 cm (0.17 inches), its small size can be advantageous during installation. Because it has such
a small external diameter, UTP does not fill up wiring ducts as rapidly as other types of cable.
This can be an extremely important factor to consider, particularly when installing a network in
an older building. UTP cable is easy to install and is less expensive than other types of
networking media. In fact, UTP costs less per meter than any other type of LAN cabling. And
because UTP can be used with most of the major networking architectures, it continues to grow
in popularity.
Disadvantages also are involved in using twisted-pair cabling, howe.
More Related Content
Similar to Heterozygous FH means a person has inhrited the fsulty gene from eit.pdf
If the characteristic followed in the pedigree is x-linked recessive .pdf
If the characteristic followed in the pedigree is x-linked recessive allele, what is III-1\'s
genotype? a. Hemizygous for a dominant allele b. Hemizygous for a recessive allele c.
Definitely heterozygous d. dominant e. Either heterozygous or homozygous dominant
Solution
1st generation parents 1,2 though are normal but must be heterozygotes (Pp) since their children
are affected i.e. 2nd generation males 1,3.
Since 1st generation male 3 is affected (pp) and 4 is normal but regardless if 4 in 1st generation
is PP or Pp their children will be either 4 Pp i.e. heterozygote or 2 affected pp and 2
heterozygotes Pp. But the 2nd generation 4, 5 female are normal but heterozygotes Pp.
So the cross between a 2nd generation affected male 3(pp) and a normal but heterozygote female
4(Pp) would have a normal male child only if hemizygous for a dominant allele condition exists.
Hemizygous is a condition in which only one gene copy/DNA sequence exists in diploid cells.
Males are hemizygous for most genes on sex chromosomes since they have only one X and one
Y chromosome unlike two X for females. A recessive X-linked gene requires two mutant alleles
to affect the female phenotype.
Rules followed here are.
Hemophilia is a recessive trait. The dominant trait is normal blood d.pdf
Hemophilia is a recessive trait. The dominant trait is normal blood dotting A gene that can cause
hemophilia is located on the X chromosome. If a homozygous normal female marries a normal
male, they children might have which of the following phenotypes: Select one: a. normal
females, hemophiliac females, and hemophiliac males. b. normal females, hemophiliac females,
normal males, and hemophiliac males. c. normal females and hemophiliac males d. normal
females and normal mates. e. normal females, normal males or hemophiliac males.
Solution
In X - linked recessive inheritance males are more prone to disease because they carry the
disease trait. If the affected offspring’s mother has brother or father containing the same trait
then the mother pass the trait to her offspring. The affected males produce unaffected daughters
who can produce 50% affected sons. Hemophilia is the example of X - linked recessive trait.
When a homozygous normal female marries to a normal male the offspring produced are normal
females and normal males. Option d is correct..
Ferdinand is affected by hemophilia, a recessive x-linked condition..pdf
Ferdinand is affected by hemophilia, a recessive x-linked condition. He mates with an unaffected
woman with no history of hemophilia in her family. Their first child is a girl with Turner
Syndrome (XO sex chromosomes), who is also hemophiliac. You infer that she resulted from the
following sperm and egg (answers listed in the order of sperm, egg):normal, normal
Solution
Answer: Because Ferdinand has hemophilia which is X-linked ressesive disease. Daughters
produced by mating of ferdinand and a normal female would be carriers of hemophilia. In this
case however, the daughter is having Turners\' syndrome (XO) and is heophilic as well. This
implies that the X came from sperm of Ferdinand and the O is beacuse the X chromosome from
the female was nondisjunct at meiosis I, as non-disjunction at meiosis II would result in two
gaetes being normal, One being trisomic and one being monosomic. whereas nondysjunction at
meiosis I would give rise to 50% trisomic and 50% monosomic gametes. Hence raising the
possibility of turners\' syndrome after nondysjunction at meiosis I.
Therefore answer is:
c. Normal and non-dysjunction at meiosis I..
Sex-linked inheritance by Puzon and Tope
Sex-linked inheritance by Puzon, kathleen Faye R. and Tope, Donita J.
Inbreeding and outbreeding
This powerpoint gives a clear picture on inbreeding and also about outbreeding of higher organisms. This also explains the advantages and disadvantages of the above said topics. the methods of inbreeding and reasons for inbreeding also given in this powerpoint.
1. In a given plant, the yellow seed color allele (Y) is dominant to .pdf
1. In a given plant, the yellow seed color allele (Y) is dominant to green seed color (y). Give the
genotypic and phenotypic ratios for the following crosses? a. homozygous yellow x green b.
heterozygous yellow x green c. heterozygous yellow x heterozygous yellow d. green x green 2. If
two animals heterozygous for a single pair of genes are mated and have 276 offspring, how many
would be expected to have the dominant genotype? 3. Two long-winged fies are mated. The
offspring included 90 with long wings and 32 with short wings. Which phenotype is recessive?
4. A blood test reveals that a couple both have the sickle-cell trat. What is the probability of their
producing a child with sickle-cell anemia? 5. A blood type A woman has a type B child. She
accuses a type O man of being the father. Knowing genetics you are asked to give an opinion; do
so. 6. Assume that human eye color is controlled by a single pair of alleles. A blue-eyed man,
both of whose parents were bom brown-eyed, marries a brown-eyed woman whose father was
blue-eyed and whose mother was brown-eyed. They have a blue-eyed child. Which are the
genotypes of all individuals mentioned? 7. A red fruit, tall tomato plant RrTt, is crossed with a
yellow-fruited dwarf plant ritt. What genotypes and phenotypes would be expected in the
offispring? 8. What genotype and phenotype would be expected in the offspring if the cross in
the previous problem had been RrTt x RrTt ?.
sex linked inheritance, Sex Influence inheritance and sex limited characters
sex linked inheritance, Sex Influence inheritance and sex limited characters
Similar to Heterozygous FH means a person has inhrited the fsulty gene from eit.pdf (16)
If the characteristic followed in the pedigree is x-linked recessive .pdf
If the characteristic followed in the pedigree is x-linked recessive .pdf
Hemophilia is a recessive trait. The dominant trait is normal blood d.pdf
Hemophilia is a recessive trait. The dominant trait is normal blood d.pdf
Ferdinand is affected by hemophilia, a recessive x-linked condition..pdf
Ferdinand is affected by hemophilia, a recessive x-linked condition..pdf
1. In a given plant, the yellow seed color allele (Y) is dominant to .pdf
1. In a given plant, the yellow seed color allele (Y) is dominant to .pdf
sex linked inheritance, Sex Influence inheritance and sex limited characters
sex linked inheritance, Sex Influence inheritance and sex limited characters
More from aquazac
Answer Every investor expects dividend from his investments.Dividen.pdf
Answer: Every investor expects dividend from his investments.Dividend income helps investors
in dealing with their routine expenses. Investors put their money to earn some positive returns
and dividend is one of the way of getting returns for investors. So, It is a duty of management to
satisfy dividend expectation of investors.
If management do not pay dividend then it is known as retention of earnings. Retention earnings
has it\'s own pro and cons, which are described as follows:
Pros:
Retained earnings reduces organization\'s dependence on external borrowings. It can be
considered as an advantage when external borrowings are not available easily.
Retained earnings indicates good business opportunities, It means organization is growing
financially. It may result in better stock valuation of organization.
In some situation dividend is subject to taxes (either distribution tax or income tax) in the hands
of investors. So, in this case investors are not interested in dividend but they are more interested
in capital gain(which may be tax exempt). So, In this situation retained earnings fulfill investor\'s
objective.
Cons:
As explained earlier all the investors have some expectation of dividend, in case of higher
retention of earnings it may become difficult to fulfill investor\'s expectations.
Retention of earnings may affect investor\'s earning capacity when investors have available
opportunities, which gives higher return than organization earns by retaining amount. In this case
investor looses earning opportunities.
In case of dividend management\'s decision is significantly influenced by media and institutional
investors. Media is responsible for organization\'s image in society, such image significantly
affects organization\'s financial performance. This is why management always considers media\'s
point of view before making any dividend decision.
Institutional investors are usually invests their money and takes part in decision making process
of organization. Institutional investor\'s weight in decision making process is generally very
high, so it is obvious that they influence management\'s dividend decision also.
Solution
Answer: Every investor expects dividend from his investments.Dividend income helps investors
in dealing with their routine expenses. Investors put their money to earn some positive returns
and dividend is one of the way of getting returns for investors. So, It is a duty of management to
satisfy dividend expectation of investors.
If management do not pay dividend then it is known as retention of earnings. Retention earnings
has it\'s own pro and cons, which are described as follows:
Pros:
Retained earnings reduces organization\'s dependence on external borrowings. It can be
considered as an advantage when external borrowings are not available easily.
Retained earnings indicates good business opportunities, It means organization is growing
financially. It may result in better stock valuation of organization.
In some situ.
Ans. Gene is defined as the segment of DNA that gives a functional p.pdf
Ans. Gene is defined as the segment of DNA that gives a functional product (protein, RNA). A
functional product can be either protein (or polypeptide) or RNA.
mRNA transcribes from a gene is further translated to produce polypeptides. However, all other
RNAs are also transcribed from their respective genes. It includes tRNA, rRNA, snRNA and
various regulatory RNAs (micro RNA, small interfering RNA), etc. tRNA acts as amino-acid
carrier during translation. rRNA is the structural and functional scaffold of ribosomes. snRNA
mediates splicing of immature mRNA. miRNA and siRNA have crucial roles in gene
regulations.
SO, polypeptides are just one class of functional product from DNA. The genomic DNA also
produces various classes of RNAs with diverse functions that are crucial for cellular activates
and gene regulation.
Therefore, the statement “Add DNA sequences in the genome that does not encode polypeptides
can be defined as junk DNA” is INCORRECT. Junk DNA is non-coding DNA that excludes all
genes (including those encoding polypeptides and RNAs).
Solution
Ans. Gene is defined as the segment of DNA that gives a functional product (protein, RNA). A
functional product can be either protein (or polypeptide) or RNA.
mRNA transcribes from a gene is further translated to produce polypeptides. However, all other
RNAs are also transcribed from their respective genes. It includes tRNA, rRNA, snRNA and
various regulatory RNAs (micro RNA, small interfering RNA), etc. tRNA acts as amino-acid
carrier during translation. rRNA is the structural and functional scaffold of ribosomes. snRNA
mediates splicing of immature mRNA. miRNA and siRNA have crucial roles in gene
regulations.
SO, polypeptides are just one class of functional product from DNA. The genomic DNA also
produces various classes of RNAs with diverse functions that are crucial for cellular activates
and gene regulation.
Therefore, the statement “Add DNA sequences in the genome that does not encode polypeptides
can be defined as junk DNA” is INCORRECT. Junk DNA is non-coding DNA that excludes all
genes (including those encoding polypeptides and RNAs)..
additional optmization techniques for underlying IP network must1.pdf
additional optmization techniques for underlying IP network must:
1.Have bandwidth and performance to handle converged services.
2.Seprate VLAN for Voice traffic.
3.Meet the demand of high availability of voice services by providing resiliency to mitigate the
effect of network outages.
4.Should be modular, hierarchical & consistent to promote consistency and manageability.
5.configure QoS on core switch.
Solution
additional optmization techniques for underlying IP network must:
1.Have bandwidth and performance to handle converged services.
2.Seprate VLAN for Voice traffic.
3.Meet the demand of high availability of voice services by providing resiliency to mitigate the
effect of network outages.
4.Should be modular, hierarchical & consistent to promote consistency and manageability.
5.configure QoS on core switch..
According to the given equation, aqueous carbon dioxide reacts with .pdf
According to the given equation, aqueous carbon dioxide reacts with water to form carbonic
acid, Carbonic acid is not stable in normal body temperatures, therefore, it ionizes into
bicarbonate ion (HCO3-) and a hydrogen ion (H+) is liberated. The liberation of hydrogen ions
decreases the pH.
During breathing, oxygen is inhaled which is carried by the blood to the different tissues of the
body. Oxygen facilitates the various metabolic reactions occurring inside our cells and helps in
respiration, or the generation of energy. Carbon dioxide, formed as a by-product of these energy-
generating metabolic reactions, diffuses out of the tissues and into the blood stream, which
carries CO2 to the lungs, where it is exhaled out.
a) In case of hypoventilation :
The exchange of gases that occurs in the lungs while breathing is impaired. Therefore, the
requisite amount of carbon dioxide is not being exhaled out. This leads to a build-up of carbon-
dioxide in the blood plasma. As a result, the forward reaction is facilitated and more H+ ions are
generated. The higher concentration of H+ ions decreases the pH (acidic pH). This can lead to
acidosis.
The build-up of carbon dioxide in the blood facilitates the forward reaction and therefore the
equilibrium shifts to the right.
The hydronium ion or H3O+ is the product of a hydrogen ion (H+) and water. Since a hydrogen
ion is too small to exist freely in water, it therefore combines with a water molecule to form a
hydronium ion which is more stable. In case of hypoventilation, the concentration of H+ ions
increase. Therefore, the concentration of hydronium ions increase during hypoventilation.
The stress concentration or partial pressure of CO2 during hypoventilation is significantly higher
than 40mmHg (around 60-65 mmHg).
b) In case of hyperventilation :
The breathing rate is more than normal. Therefore, as a result of heavier and faster breathing, the
CO2 concentration in the blood rapidly decreases, that is, CO2 easily leaves the blood plasma.
Decrease in the CO2 concentration slows down the rate of the forward reaction, reducing the
levels at which bicarbonate ion and H+ ion concentration were being generated. The bicarbonate
and hydrogen ions recombine to yield CO2 and water. This lowering in hydrogen ion
concentration consequently lowers the pH (alkaline pH). This can lead to alkalosis.
As a result of the lowering of CO2 , the reverse reaction takes place. The carbonic acid changes
back to CO2 and H2O, shifting the equilibrium to the left.
The concentration of H3O+ decreases during hyperventilation, since the H+ ion concentration is
lowered.
The stress levels or partial pressure of CO2 during hyperventilation is significantly lower than 40
mmHg (around 25-30 mmHg).
c) If sodium carbonate or Na2CO3 is added instead of sodium bicarbonate, then sodium
carbonate would react with water and it would hydrolyse into Na+ and CO32- ions. The CO32-
combine with the hydrogen (H) present in water to form H2CO3 (carbon.
a) mean = 1.43Thus distribution is Poisson(4.2)P(X = 4) = 4.2^4.pdf
a) mean = 1.4*3
Thus distribution is Poisson(4.2)
P(X = 4) = 4.2^4 * e^{-4.2} / 4! = 0.1944
b)
average number of ghosts in 90min (1.5hrs) interval = 1.4*1.5 = 2.1
c)
Probability that 4th ghost will appear before 10 am
= Probability that number of ghosts appearing from 8 to 10 am >= 4
As it is a 2 hr period, the distribution of number of ghosts is Poisson(1.4*2) i.e Poisson(2.8)
P(X >= 4)
= 1 - P(X < 4)
= 1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3)]
= 1 - [(2.8^0 * e^{-2.8} / 0!) + (2.8^1 * e^{-2.8} / 1!) + (2.8^2 * e^{-2.8} / 2!) + (2.8^3 * e^{-
2.8} / 3!)]
= 1 - [0.0608 + 0.1703 + 0.2384 + 0.2225]
= 0.692
d)
When arrival distribution is Poisson(lambda), inter-arrival time distribution is
Exponential(lambda)
Thus, interarrival time distribution is Exponential(1.4)
Expected value of time between 5th and 6th ghost = 1/1.4 = 0.7143hr
e)
On average time taken for 10th ghost to arrive = 10*(1/1.4) = 7.143hr
Thus on an average, 10th ghost will arrive at 3:09pm
f)
Expected value of time between ghost appearances = 1/1.4 = 0.7143hr (see part d)
g)
Probability that 3rd ghost appears after 9:45 am
= Probablity that number of ghosts appearing till 9:45 am < 3
Duration from 8am to 9:45 am = 1.75hr
Mean number of ghosts = 1.4*1.75 = 2.45
Thus X is Poisson(2.45)
P(X < 3)
= P(X = 0) + P(X = 1) + P(X = 2)
= (2.45^0 * e^{-2.45} / 0!) + (2.45^1 * e^{-2.45} / 1!) + (2.45^2 * e^{-2.45} / 2!)
= 0.0863 + 0.2114 + 0.2590
= 0.5567
h)
As inter-arrival time is Exponential(1.4), inter-arrival time is memoryless
Thus, time for 7th ghost will be calculated from 1:00 pm and not from 12:35pm due to
memorylessness property.
Time duration from 1:00pm to 1:15pm = 0.25hrs
Let Y be random variable denoting inter-arrival time
probability that the 7th ghost will appear before 1:15 p.m.
= P(Y < 0.25)
= 1 - e^{-1.4 * 0.25)
= 0.2953
i)
Expected value of time for 7th ghost to appear after 1:00pm = 1/1.4 = 0.7143hr = 42.86 min
Time from 12:35pm to 1:00pm = 25min
Thus expected interarrival time = 25 + 42.86 = 67.86 min = 1hr 7.86min
j)
Expected duration after 1:00pm for 9th ghost to appear = 3 * (1/1.4) = 2.143hr = 2hr 8.58min
Thus expected time at which 9th ghost appears = 3:09pm
k)
As the inter-arrival time follows memorylessness property, the future probabilities do not change
with the observed value at past.
Thus, 4 ghosts appearing between 7:00pm to 10:00pm does not affect ghost appearance from
11:00pm to 11:30pm
Duration from 11:00pm to 11:30pm = 0.5hrs
mean number of ghosts = 1.4*0.5 = 0.7
Thus it follows Poisson(0.7) distribution
P(X > 2)
= 1 - P(X <= 2)
= 1 - [P(X = 0) + P(X = 1) + P(X = 2)]
= 1 - [(0.7^0 * e^{-0.7} / 0!) + (0.7^1 * e^{-0.7} / 1!) + (0.7^2 * e^{-0.7} / 2!)]
= 1 - [0.4966 + 0.3476 + 0.1217]
= 0.0341
l)
Probability that 8 ghosts appear between 6pm and midnight
= Probability that number of ghosts appear from 6pm to 7pm and 10pm to midnight = 4 as it is
known that 4 ghosts appeared from 7pm to 10pm
Let A denote number of ghost from 6pm to 7pm
B denote number of ghosts .
2.a. Wired Media Type and ExplinationTwisted-Pair CableTwiste.pdf
2.
a. Wired Media Type and Explination:
Twisted-Pair Cable
Twisted-pair cable is a type of cabling that is used for telephone communications and most
modern Ethernet networks. A pair of wires forms a circuit that can transmit data. The pairs are
twisted to provide protection against crosstalk, the noise generated by adjacent pairs. When
electrical current flows through a wire, it creates a small, circular magnetic field around the wire.
When two wires in an electrical circuit are placed close together, their magnetic fields are the
exact opposite of each other. Thus, the two magnetic fields cancel each other out. They also
cancel out any outside magnetic fields. Twisting the wires can enhance this cancellation effect.
Using cancellation together with twisting the wires, cable designers can effectively provide self-
shielding for wire pairs within the network media.
Two basic types of twisted-pair cable exist: unshielded twisted pair (UTP) and shielded twisted
pair (STP). The following sections discuss UTP and STP cable in more detail.
UTP Cable
UTP cable is a medium that is composed of pairs of wires . UTP cable is used in a variety of
networks. Each of the eight individual copper wires in UTP cable \\is covered by an insulating
material. In addition, the wires in each pair are twisted around each other.
UTP cable relies solely on the cancellation effect produced by the twisted wire pairs to limit
signal degradation caused by electromagnetic interference (EMI) and radio frequency
interference (RFI). To further reduce crosstalk between the pairs in UTP cable, the number of
twists in the wire pairs varies. UTP cable must follow precise specifications governing how
many twists or braids are permitted per meter (3.28 feet) of cable.
UTP cable often is installed using a Registered Jack 45 (RJ-45) connector The RJ-45 is an eight-
wire connector used commonly to connect computers onto a local-area network (LAN),
especially Ethernets.
When used as a networking medium, UTP cable has four pairs of either 22- or 24-gauge copper
wire. UTP used as a networking medium has an impedance of 100 ohms; this differentiates it
from other types of twisted-pair wiring such as that used for telephone wiring, which has
impedance of 600 ohms.
UTP cable offers many advantages. Because UTP has an external diameter of approximately
0.43 cm (0.17 inches), its small size can be advantageous during installation. Because it has such
a small external diameter, UTP does not fill up wiring ducts as rapidly as other types of cable.
This can be an extremely important factor to consider, particularly when installing a network in
an older building. UTP cable is easy to install and is less expensive than other types of
networking media. In fact, UTP costs less per meter than any other type of LAN cabling. And
because UTP can be used with most of the major networking architectures, it continues to grow
in popularity.
Disadvantages also are involved in using twisted-pair cabling, howe.
A person may not choose to participate in the labour force due to La.pdf
A person may not choose to participate in the labour force due to Laziness, lack of ambition,
emotional or chemical imbalance, independent wealth, personal unavailability, or other priorities.
However, race, location, education, and marital status are irrelevant at times. They do affect the
type of job, or their availability, but not the individual\'s desire to have one. Choosing to stay
home and raise children is an example of different priorities, and a spouse\'s income, if they are
wealthy enough, may negate having to even make a choice.
(a) with low education, a job might not pay much more than unemployment benefit.
(b) with children, there is strong incentive to stay at home and take care of them.
(c) with one spouse earning a lot, another spouse can stay at home.
(d) if person belongs to a race with bad reputation, they are less likely to get hired. But at times
women are viewed as better workers than men from same race.
(e) in rural areas, there are fewer jobs, and the commute to them is longer and therefore they
prefer to do some small work nearby or choose not to work at all.
Solution
A person may not choose to participate in the labour force due to Laziness, lack of ambition,
emotional or chemical imbalance, independent wealth, personal unavailability, or other priorities.
However, race, location, education, and marital status are irrelevant at times. They do affect the
type of job, or their availability, but not the individual\'s desire to have one. Choosing to stay
home and raise children is an example of different priorities, and a spouse\'s income, if they are
wealthy enough, may negate having to even make a choice.
(a) with low education, a job might not pay much more than unemployment benefit.
(b) with children, there is strong incentive to stay at home and take care of them.
(c) with one spouse earning a lot, another spouse can stay at home.
(d) if person belongs to a race with bad reputation, they are less likely to get hired. But at times
women are viewed as better workers than men from same race.
(e) in rural areas, there are fewer jobs, and the commute to them is longer and therefore they
prefer to do some small work nearby or choose not to work at all..
Well.. 1) Ionic bonds are almost always metal to .pdf
Well.. 1) Ionic bonds are almost always metal to non-metal. (opposing charges
attract). 2) Covalent, electron-sharing. \"co\"-valent. Valent referring to the valence electrons of
the element, that is, the electrons in outer most shell. Remember each element wants to reach a
more stable state. To achieve this it will share electrons with another element through covalent
bonds. 3) a covalent bond is said to be polar if one of the elements sharing the electrons has a
greater electronegativity. (i.e. water contains polar covalent bonds between the oxygen and the
hydrogens because oxygen is far more electronegative than hydrogen, therefore it \"hogs\" the
electrons. 4) non-polar covalent bonds have an equal sharing of the electrons, equal
electronegativity. Hope this helps!
Solution
Well.. 1) Ionic bonds are almost always metal to non-metal. (opposing charges
attract). 2) Covalent, electron-sharing. \"co\"-valent. Valent referring to the valence electrons of
the element, that is, the electrons in outer most shell. Remember each element wants to reach a
more stable state. To achieve this it will share electrons with another element through covalent
bonds. 3) a covalent bond is said to be polar if one of the elements sharing the electrons has a
greater electronegativity. (i.e. water contains polar covalent bonds between the oxygen and the
hydrogens because oxygen is far more electronegative than hydrogen, therefore it \"hogs\" the
electrons. 4) non-polar covalent bonds have an equal sharing of the electrons, equal
electronegativity. Hope this helps!.
1. The answer is d) Environmental EffectsEnvironmental effects ca.pdf
1. The answer is: d) Environmental Effects
Environmental effects can cause a single genotype to produce a range of potential Phenotypes.
2. The answer is: c) Differ in both mean and variance
The Figure represents two distributions that differ in both mean and variance.
Solution
1. The answer is: d) Environmental Effects
Environmental effects can cause a single genotype to produce a range of potential Phenotypes.
2. The answer is: c) Differ in both mean and variance
The Figure represents two distributions that differ in both mean and variance..
clear clc close all Use polyfit to solve for the phase l.pdf
clear
clc
close all
%Use polyfit to solve for the phase lines
point11=[0,15];
point12=[0,300];
line1=polyfit(point11,point12,1);
m1=line1(1);
b1=line1(2);
xpf1=linspace(min(point11),max(point11),100);
ypf1=m1*xpf1+b1;
point21=[0,15];
point22=[700,300];
line2=polyfit(point21,point22,1);
m2=line2(1);
b2=line2(2);
xpf2=linspace(min(point21),max(point21),100);
ypf2=m2*xpf2+b2;
point31=[0,50];
point32=[700,300];
line3=polyfit(point31,point32,1);
m3=line3(1);
b3=line3(2);
xpf3=linspace(min(point31),max(point31),100);
ypf3=m3*xpf3+b3;
point41=[50,100];
point42=[300,800];
line4=polyfit(point41,point42,1);
m4=line4(1);
b4=line4(2);
xpf4=linspace(min(point41),max(point41),100);
ypf4=m4*xpf4+b4;
point51=[100,85];
point52=[800,300];
line5=polyfit(point51,point52,1);
m5=line5(1);
b5=line5(2);
xpf5=linspace(min(point51),max(point51),100);
ypf5=m5*xpf5+b5;
point61=[85,100];
point62=[300,0];
line6=polyfit(point61,point62,1);
m6=line6(1);
b6=line6(2);
xpf6=linspace(min(point61),max(point61),100);
ypf6=m6*xpf6+b6;
point71=[15,85];
point72=[300,300];
%Allow the user to set the percent of B
b=input(\'Enter the mass percent of B: \');
%Return an error if the percent of B is not a valid number
if b<0 || b>100
error(\'Please enter a value between 0 and 100.\')
end
%Allow the user to set the temperature
t=input(\'Enter the temperature [deg C]: \');
%Solve for percent composition of A
a=100-b;
%Determine phase based on temperature and percent composition of B using
%the phase lines
if t<300
if b<15 && t>m1*b+b1
phase=\'alpha\';
elseif b<15 && t=15 && b<=85
phase=\'alpha + beta\';
elseif b>85 && t>m6*b+b6
phase=\'beta\';
else phase=\'alpha + beta\';
end
elseif b<15
if t=15 && b<=50
if t50 && b<=85
if t=15 && b<=85 && t==300
fprintf(\'The provided conditions fall on the eutectic line.\ \')
end
if b==50 && t==300
fprintf(\'The provided conditions fall on the eutectic point.\ \')
end
%Output a formatted statement to display the phase
fprintf(\'For the composition of %0.2f%% A, %0.2f%% B and a temperature of %0.0f degrees
Celsius, the phase is %s.\ \',a,b,t,phase)
%Create a plot of the phase lines and the user\'s set point
figure(\'color\',\'white\')
%Set axes
axis([0 100 0 1000])
%Plot the phase lines, eutectic line, eutectic point, and the user\'s point
plot(xpf1,ypf1,\'-b\')
hold on
plot(xpf2,ypf2,\'-b\')
plot(xpf3,ypf3,\'-b\')
plot(xpf4,ypf4,\'-b\')
plot(xpf5,ypf5,\'-b\')
plot(xpf6,ypf6,\'-b\')
plot(point71,point72,\'-k\')
plot(b,t,\'ro\',\'MarkerFaceColor\',\'r\')
%Assign a title and axes labels
title(\'Phase Diagram of Elements A and B\')
xlabel(\'Percent Composition of B\')
ylabel(\'Temperature (T) [deg C]\')
%Set scaling for the axes
set(gca,\'XTick\',0:10:100,\'YTick\',0:100:1000)
%Place the phase names on the graph
text(40,200,\'alpha+beta\')
text(45,600,\'Liquid\')
text(90,300,\'beta\')
text(3,300,\'alpha\')
text(15,375,\'alpha + Liquid\')
text(62,375,\'beta + Liquid\')
Solution
clear
clc
close all
%Use polyfit to solve for the phase lines
point11=[0,15];
point.
The oxygen appears in both step reactions. But, i.pdf
The oxygen appears in both step reactions. But, it would not appear in the final
reaction. Therefore it is a reaction intermediate. It is sort of in the \"middle\" The final reaction
is 2O3 ====> 3 O2 (No O)
Solution
The oxygen appears in both step reactions. But, it would not appear in the final
reaction. Therefore it is a reaction intermediate. It is sort of in the \"middle\" The final reaction
is 2O3 ====> 3 O2 (No O).
PART A The element Si belongs to IVA group. Therefore, four electro.pdf
PART A: The element Si belongs to IVA group. Therefore, four electrons present in the
outermost shell. Hence the condensed electron configuration of Si is [Ne] 3s23p2. PART B: The
s orbital is completely filled and there are two electrons present in p-orbital. Therefore, the
number of unpaired electrons of Si is 2 PART C: The atomic number of Zn is 30 and it belongs
to IIB group. Therefore, the condensed electron configuration of Zn is [Ar] 4s23d10. PART D:
All orbitals of Zn completely filled with electrons. Therefore, there are no unpaired electrons.
PART E: The atomic number of Zr is 40 and it belongs to IVB group. This element is present
after third element of Kr in periodic table. Hence the condensed electron configuration of Zr is
[Kr] 5s24d2. PART F: The s orbital is completely with electrons and there are two electrons
present in the d orbital. Therefore, the number of unpaired electrons of Zr is 2 PART G: The
element Sn belongs to IVA group. Therefore, four electrons present in the outermost shell.
Hence the condensed electron configuration of Sn is [Kr] 5s24d10 5p2. PART H: The s orbital
and p orbital is completely filled. Therefore, there are two unpaired electrons in the Sn metal.
PART I: The element Ba belongs to IIA group. Therefore, two electrons present in the s-orbital.
Hence the condensed electron configuration of Ba is [Xe] 6s2. PART J: The outermost s-
electron is completely filled. Therefore, there are no unpaired electrons present in Ba metal.
PART K: The element Tl belongs to IIIA group. Therefore, the condensed electron configuration
of Tl is [Xe] 4f14 5d10 6s2 6p1. PART L: The all orbitals are filled with electrons except Tl.
Therefore, the number of unpaired electrons is 1
Solution
PART A: The element Si belongs to IVA group. Therefore, four electrons present in the
outermost shell. Hence the condensed electron configuration of Si is [Ne] 3s23p2. PART B: The
s orbital is completely filled and there are two electrons present in p-orbital. Therefore, the
number of unpaired electrons of Si is 2 PART C: The atomic number of Zn is 30 and it belongs
to IIB group. Therefore, the condensed electron configuration of Zn is [Ar] 4s23d10. PART D:
All orbitals of Zn completely filled with electrons. Therefore, there are no unpaired electrons.
PART E: The atomic number of Zr is 40 and it belongs to IVB group. This element is present
after third element of Kr in periodic table. Hence the condensed electron configuration of Zr is
[Kr] 5s24d2. PART F: The s orbital is completely with electrons and there are two electrons
present in the d orbital. Therefore, the number of unpaired electrons of Zr is 2 PART G: The
element Sn belongs to IVA group. Therefore, four electrons present in the outermost shell.
Hence the condensed electron configuration of Sn is [Kr] 5s24d10 5p2. PART H: The s orbital
and p orbital is completely filled. Therefore, there are two unpaired electrons in the Sn metal.
PART I: The element Ba belongs to IIA group. Th.
The two contributions to the cohesive energy of t.pdf
The two contributions to the cohesive energy of the ionic crystal are : Vander waals
attraction between the adjacent ions Zero point oscillations of their equilibrium position. Zero-
point energy is the lowest possible energy that a quantum mechanical physical system may have;
it is the energy of its ground state. All quantum mechanical systems undergo fluctuations even in
their ground state and have an associated zero-point energy, a consequence of their wave-like
interaction.
Solution
The two contributions to the cohesive energy of the ionic crystal are : Vander waals
attraction between the adjacent ions Zero point oscillations of their equilibrium position. Zero-
point energy is the lowest possible energy that a quantum mechanical physical system may have;
it is the energy of its ground state. All quantum mechanical systems undergo fluctuations even in
their ground state and have an associated zero-point energy, a consequence of their wave-like
interaction..
If you are talking about an extraction design, th.pdf
If you are talking about an extraction design, the the water solubility has to be
greater in the aqueous base because of the greater solubility of the salts of the acids (with few
exceptions), again presuming a differential extraction aqueous to organic.
Solution
If you are talking about an extraction design, the the water solubility has to be
greater in the aqueous base because of the greater solubility of the salts of the acids (with few
exceptions), again presuming a differential extraction aqueous to organic..
Yes ,its true. Though both gibbons and rhesus monkeys belong to pr.pdf
Yes ,it\'s true. Though both gibbons and rhesus monkeys belong to primates, but gibbons are
under a group apes, and rhesus monkeys belongs to old world monkey group. Apes are near
common ancestor of homosapiens.
Solution
Yes ,it\'s true. Though both gibbons and rhesus monkeys belong to primates, but gibbons are
under a group apes, and rhesus monkeys belongs to old world monkey group. Apes are near
common ancestor of homosapiens..
When something boils, it changes states of matter. It would go from .pdf
When something boils, it changes states of matter. It would go from a liquid to a soild when it
boils, so if more heat is added, it will just change the state of matter faster.
Solution
When something boils, it changes states of matter. It would go from a liquid to a soild when it
boils, so if more heat is added, it will just change the state of matter faster..
We need to discuss why there is an importance of adding residents to.pdf
We need to discuss why there is an importance of adding residents to the budget planning, we
can go through below explaintion and can derived various conclusions from the same.
In explaining why they did or did not seek input from citizens, respondents in- dicated that the
process was guided by the desires of the leadership. Of those who said that they did seek citizen
input, the desire (as opposed to the obligation through policy) of the staff or the govern- ing
board or staff to seek public opinion seemed to provide some motivation. Almost three-quarters
of responding cities cited staff desire, and almost two-thirds, governing board desire.
The pattern for counties was similar. The most often cited reason for extra effort to involve the
public was board interest. Almost half of the counties also cited staff interest.
Although citizen involvement is a hot topic in management literature these days, local
governments in North Caro- lina are generally not making extensive fforts to involve residents in
budgeting. Barely half of the cities responding to the survey and less than a third of the coun- ties
go beyond the single, mandated hearing to get citizen input In general, cities tend to seek input
more than counties do. This result might be explained by the different types of services offered
by the two forms of gov- ernment. Counties provide funding for, among other things, social
services, pub- lic health services, mental health ser- vices, and schools, and the boards over-
seeing these services may seek citizen input when they are preparing their own budget request
for the county.
Although there is informal interest in hearing from citizens, boards clearly want to remain
flexible in when and how they obtain input. Only 13 percent of the city respondents had a formal
pol- icy or requirement for citizen involve- ment beyond the single, state-mandated hearing.
Medium to large cities were more likely to have such requirements than small cities. Only one
county, a larger one, reported having such a policy or requirement.
Although citizen involvement is a hot topic in management literature these days, local
governments in North Caro- lina are generally not making extensive
efforts to involve residents in budgeting. Barely half of the cities responding to the survey and
less than a third of the coun- ties go beyond the single, mandated hearing to get citizen input
Of those that did not seek extra citizen involvement, the most common reason cited, on both the
city and the county level, was lack of governing board inter- est. The second most common
reason was poor response in the past, when at- tempts to get citizens involved did not seem very
effective. At the county level, lack of board interest was a far more important factor than
anything else, including poor past response: 71 percent of respondents cited lack of board
interest, whereas only 39 percent cited poor past response Although staff interest seemed to be a
big motivator for involving ci.
What is the largest decimal integer that can be represented with the.pdf
What is the largest decimal integer that can be represented with the following number of bits?
1. 3 bits
2. 8 bits
3. 32 bits
4. 64 bits
How do you find out the largest decimal integer?
Solution
In general, with k bits we can represent 2k values. Their range will depend on the system:
Unsigned: 0 to 2k-1
1. 3 bits
=> 2^3 - 1 => 7 => 111
2. 8 bits
=> 2^8 - 1 => 255
3. 32 bits
=> 2^32 - 1
4. 64 bits
=> 2^64 - 1.
Throwing.javaimport java.util.InputMismatchException; import jav.pdf
Throwing.java
import java.util.InputMismatchException;
import java.util.Scanner;
public class Throwing
{
public static void main(String[] args)
{
try{
int n = getInt();
System.out.println(n);
}
catch(IllegalArgumentException e){
System.out.println(e.getMessage());
}
}
public static int getInt()
{
// your code goes here
int n;
try{
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter an integer: \");
n = scan.nextInt();
}
catch(InputMismatchException e){
throw new IllegalArgumentException(\"Invalid Input. Please enter an integer.\");
}
return n;
}
}
Output:
Enter an integer:
s
Invalid Input. Please enter an integer.
Throwing.java
import java.io.IOException;
import java.util.InputMismatchException;
import java.util.Scanner;
public class Throwing
{
public static void main(String[] args)
{
try{
int n = getInt();
System.out.println(n);
}
catch(IOException e){
System.out.println(e.getMessage());
}
}
public static int getInt() throws IOException
{
// your code goes here
int n;
try{
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter an integer: \");
n = scan.nextInt();
}
catch(InputMismatchException e){
throw new IOException(\"Invalid Input. Please enter an integer.\");
}
return n;
}
}
Output:
Enter an integer:
d
Invalid Input. Please enter an integer.
Solution
Throwing.java
import java.util.InputMismatchException;
import java.util.Scanner;
public class Throwing
{
public static void main(String[] args)
{
try{
int n = getInt();
System.out.println(n);
}
catch(IllegalArgumentException e){
System.out.println(e.getMessage());
}
}
public static int getInt()
{
// your code goes here
int n;
try{
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter an integer: \");
n = scan.nextInt();
}
catch(InputMismatchException e){
throw new IllegalArgumentException(\"Invalid Input. Please enter an integer.\");
}
return n;
}
}
Output:
Enter an integer:
s
Invalid Input. Please enter an integer.
Throwing.java
import java.io.IOException;
import java.util.InputMismatchException;
import java.util.Scanner;
public class Throwing
{
public static void main(String[] args)
{
try{
int n = getInt();
System.out.println(n);
}
catch(IOException e){
System.out.println(e.getMessage());
}
}
public static int getInt() throws IOException
{
// your code goes here
int n;
try{
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter an integer: \");
n = scan.nextInt();
}
catch(InputMismatchException e){
throw new IOException(\"Invalid Input. Please enter an integer.\");
}
return n;
}
}
Output:
Enter an integer:
d
Invalid Input. Please enter an integer..
This electron transport is accompanied by the protons transfer into .pdf
This electron transport is accompanied by the protons transfer into intermembrane space.
Protons release potential which drives the phosphorylation of ADP to ATP.
Protons return to the matrix through a proton specific channel
Solution
This electron transport is accompanied by the protons transfer into intermembrane space.
Protons release potential which drives the phosphorylation of ADP to ATP.
Protons return to the matrix through a proton specific channel.
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According to the given equation, aqueous carbon dioxide reacts with .pdf
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a) mean = 1.43Thus distribution is Poisson(4.2)P(X = 4) = 4.2^4.pdf
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2.a. Wired Media Type and ExplinationTwisted-Pair CableTwiste.pdf
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A person may not choose to participate in the labour force due to La.pdf
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Well.. 1) Ionic bonds are almost always metal to .pdf
Well.. 1) Ionic bonds are almost always metal to .pdf
1. The answer is d) Environmental EffectsEnvironmental effects ca.pdf
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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.
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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.
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Event Link:-
https://meetups.mulesoft.com/events/details/mulesoft-mysore-presents-exploring-gemini-ai-and-integration-with-mulesoft/
Agenda
● Java 17 Upgrade Overview
● Why and by when do customers need to upgrade to Java 17?
● Is there any immediate impact to upgrading to Mule Runtime 4.6 and beyond?
● Which MuleSoft products are in scope?
For Upcoming Meetups Join Mysore Meetup Group - https://meetups.mulesoft.com/mysore/
YouTube:- youtube.com/@mulesoftmysore
Mysore WhatsApp group:- https://chat.whatsapp.com/EhqtHtCC75vCAX7gaO842N
Speaker:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Priya Shaw - https://www.linkedin.com/in/priya-shaw
Organizers:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Giridhar Meka - https://www.linkedin.com/in/giridharmeka
Priya Shaw - https://www.linkedin.com/in/priya-shaw
Shyam Raj Prasad-
https://www.linkedin.com/in/shyam-raj-prasad/
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The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
For more information, visit-www.vavaclasses.com
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- 1. Heterozygous FH means a person has inhrited the fsulty gene from either of his/her parents. Homozygous FH-condition where a person inherits two faulty oraltered grnes from both the parents. If the two faulty genes are identical it is refereed to as homozygous FH. By calculating the chance by Punnett square- X+-faulty gene 2 children -Normal FH by inheriting one faulty and one healthy gene-XX+ and XY+ 1Child-Homozygous by receiving faulty gene from each of the parents-X+X+ 1Child-Normal-i.e. will not inherit FH as the child will receive the norml genes from the parents.XY Therefore, in one family of four chldren the ratio of homozygous FH will be 1:4, it will bw about 1 in a million.X+YX+X+X+X+YXXX+XY Solution Heterozygous FH means a person has inhrited the fsulty gene from either of his/her parents. Homozygous FH-condition where a person inherits two faulty oraltered grnes from both the parents. If the two faulty genes are identical it is refereed to as homozygous FH. By calculating the chance by Punnett square- X+-faulty gene 2 children -Normal FH by inheriting one faulty and one healthy gene-XX+ and XY+ 1Child-Homozygous by receiving faulty gene from each of the parents-X+X+ 1Child-Normal-i.e. will not inherit FH as the child will receive the norml genes from the parents.XY Therefore, in one family of four chldren the ratio of homozygous FH will be 1:4, it will bw about 1 in a million.X+YX+X+X+X+YXXX+XY