Ferdinand is affected by hemophilia, a recessive x-linked condition..pdf
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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..
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The non-disjunctioning had occur in her mother.Turners syndrome .pdf
The non-disjunctioning had occur in her mother.
Turner\'s syndrome is caused by incomplete inheritance of or missing X chromosome, which
generally affects females. The poor growth and development of sexual organs can be seen early.
Haemophilia is an X-lined recessive disease. Females inherit one X-chromosome from the father
and another from the mother. The inheritance of haemophilia from the father ensures the
inheritance of X-chromosomes from the father. So, the nondisjunction occurred in females
during meiosis-I. The daughter received one X-chromosome from the father and no
chromosomes from the mother, resulting in 45, X karyotype (Turner\'s syndrome) and
haemophilia.
Solution
The non-disjunctioning had occur in her mother.
Turner\'s syndrome is caused by incomplete inheritance of or missing X chromosome, which
generally affects females. The poor growth and development of sexual organs can be seen early.
Haemophilia is an X-lined recessive disease. Females inherit one X-chromosome from the father
and another from the mother. The inheritance of haemophilia from the father ensures the
inheritance of X-chromosomes from the father. So, the nondisjunction occurred in females
during meiosis-I. The daughter received one X-chromosome from the father and no
chromosomes from the mother, resulting in 45, X karyotype (Turner\'s syndrome) and
haemophilia..
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..
Heritable Traits in Man, Pedigree Analysis and Pedigree Application
This presentation is required in our Genetics Subject
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.
Recommended
The non-disjunctioning had occur in her mother.Turners syndrome .pdf
The non-disjunctioning had occur in her mother.
Turner\'s syndrome is caused by incomplete inheritance of or missing X chromosome, which
generally affects females. The poor growth and development of sexual organs can be seen early.
Haemophilia is an X-lined recessive disease. Females inherit one X-chromosome from the father
and another from the mother. The inheritance of haemophilia from the father ensures the
inheritance of X-chromosomes from the father. So, the nondisjunction occurred in females
during meiosis-I. The daughter received one X-chromosome from the father and no
chromosomes from the mother, resulting in 45, X karyotype (Turner\'s syndrome) and
haemophilia.
Solution
The non-disjunctioning had occur in her mother.
Turner\'s syndrome is caused by incomplete inheritance of or missing X chromosome, which
generally affects females. The poor growth and development of sexual organs can be seen early.
Haemophilia is an X-lined recessive disease. Females inherit one X-chromosome from the father
and another from the mother. The inheritance of haemophilia from the father ensures the
inheritance of X-chromosomes from the father. So, the nondisjunction occurred in females
during meiosis-I. The daughter received one X-chromosome from the father and no
chromosomes from the mother, resulting in 45, X karyotype (Turner\'s syndrome) and
haemophilia..
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..
Heritable Traits in Man, Pedigree Analysis and Pedigree Application
This presentation is required in our Genetics Subject
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.
uetator Periodic Table Question 6 of 30 Incorrect Incorrect DA Saplin.pdf
uetator Periodic Table Question 6 of 30 Incorrect Incorrect DA Sapling Learning sponsible
syndrome is anX-inked, recessive condition caused by mutation of the HPRT1 muscle
movements, The syndrome is characterized byjuvenile-onset gout, in and behavioral disorders. A
mother and father have a young daughter who was diagnosed with Lesch-Nyhan syndrome. No
one in the with family was ever diagnosed the The maternal grandfather was diagnosed
syndrome, but neither one of her parents was diagnosed with condition. Select all the
descriptions that could explain how their daughter inherited Lesch-Nyhan syndrome. The
daughter could not have Nyhan because her maternal grandparents did not have disease. The girl
must have acquired the disease through a random mutation. The daughter inherited two
xchromosomes with the of the mothers X mutation. Nondisjunction chromosome chromosomes
occurred during meiosis, and the daughter inherited from her father. The father did not contribute
a sex chromosome to his daughter due to nondisjunction of the sex chromosomes. The daughter
is xo and her only X chromosome came from her mother, who was a O The d from her father,
who was a carrier. camies the Lesch-Nyhan mutation and was Her inherited There is a hint
available! bottom divider bar. Click on the divider bar again to hide the hint. Previous ® Give
Up & View solution Check Answer 0 Next HExi Hint
Solution
1) the hint here is that the mother\'s maternal was diagnosed with Lesch-Nyhan syndrome. Since
it is an X-chromosome related disease, it is very rare to have females affected by it. If a girl has
LNS, she must have inherited two effected X chromosomes from each parent. Also, one third of
cases arise from new mutations even when there is no family history of any. So the first two
points are the possible reasons for the girls to have LNS..
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..
sex linked inheritance, Sex Influence inheritance and sex limited characters
sex linked inheritance, Sex Influence inheritance and sex limited characters
Sex-linked inheritance by Puzon and Tope
Sex-linked inheritance by Puzon, kathleen Faye R. and Tope, Donita J.
Sex Linkage-Unique Heritage
This presentation intends to explore the sex-linked characters along with some fatal diseases of human beings, their cause, consequences and other issues.
In Drosophila melanogaster, the yellow body gene is X-linked. The yel.pdf
In Drosophila melanogaster, the yellow body gene is X-linked. The yellow body phenotype is
recessive to brown body phenotype. A yellow body female fruit fly was mated to a brown body
male. The vast majority of the male progeny are yellow body. However, a rare brown body male
was found in the progeny (1 out of 2060 males). What is the likely karyotype of this brown body
son? XXY XX XYY XO XXX
Solution
Sex inheritance by modified Mendelian mechanism:
X-linked inheritance means that the gene causing the trait or the disorder is located on the X
chromosome.
X-linked recessive inheritance is a mode of inheritance in which a mutation in a gene on the X
chromosome causes the phenotype to be expressed in males who are necessarily hemizygous for
the gene mutation because they have one X and one Y chromosome. Carrier females who have
only one copy of the mutation do not usually express the phenotype,
Females are XX (homogametic)
Males are XY (heterogametic)
Body color inherited by single gene, two allele system, brown dominant to yellow
But, sex and body color not inherited independently
Body color is linked to sex
Body color gene on X chromosome
Therefore, males have only one body color allele: Haploid and females have two eye color
alleles: Diploid.
Normally all the sons and none of the daughters show the recessive sex-linked characters of the
mother when the father carries the dominant allelomorph. For example, a yellow body female
mated to a brown body male produces yellow body males and brown body females, but rarely
also a yellow body female or a brown body son. The production of these exceptions by a normal
XX female must be due to an aberrant reduction division at which the two X chromosomes fail
to disjoin from each other. In consequence both remain in the egg or both pass into the polar
body. In the latter case an egg without an X chromosome is produced. Such an egg fertilized by
an X sperm produces a male with the constitution XO. These males received their single X from
their father and therefore show the father\'s characters. But that the Y does play some positive
role is proved by the fact that all the XO males have been found to be absolutely sterile.
Answer is d) XO.
A wild-type chromosome has the following segments A B C middot D E F.pdf
A wild-type chromosome has the following segments: A B C middot D E F G H I An individual
is heterozygous for the following chromosome mutation: A B C middot D H I How would the
mutated and wild-type chromosomes pair in prophase I of meiosis? D E F G H of the wild-type
chromosome will form a deletion loop. E F G of the wild-type chromosome will form a deletion
loop. D E F G H of the wild-type chromosome will form an Inversion loop. E F G of the wild-
type chromosome will form an inversion loop. Red-green color blindness is a human x-linked
recessive disorder. A young man with a 47, XXY karyotype (Klinefelter syndrome) is color
Wind. His 46, XY brother also is color blind. Both parents have normal color vision. Where did
the no disjunction that gave rise to the young man with Klinefelter syndrome take place? Assume
that no crossing over took place In prophase I of meiosis. Mother - meiosis I Mother - meiosis II
Father - meiosis I Father - meiosis II Both parents Functional epidermal lies bull nose (JEB) Is a
severe skin disorder that results in blisters over the entire body. The disorder is caused by
autosomal recessive mutations at any one of three loci that help to encode lamina 5, a major
component In the dermal-epidermal basement membrane. Leena Pulicine and colleagues
described a male newborn who was born with JEB and died at 2 months of age (L. Pulkkinen et
al. 1997. American Journal of Human Genetics 61:611-619); the child had healthy unrelated
parents. Chromosome analysis revealed that the infant had 46 normal-appearing chromosomes.
Analysis of DNA showed that his mother was heterozygous for a JEB-causing allele at the
LAMB3 locus, which is on chromosome 1. The father had two normal alleles at this locus. DNA
fingerprinting demonstrated that the male assumed to be the father had, In fact, conceived the
child. Assuming that no new mutations occurred in this family, what might best explain the
presence of an autosomal recessive disease in the child when the mother is heterozygous and the
father Is homozygous normal? The infant inherited both copies of chromosome 1 from the
mother, which is an example of unit parental disomic. The infant is monatomic for chromosome
1, which he received from his mother but did not receive a chromosome 1 from his father.
Epitasis masked the disease phenotype in the father, but complementation of the parental alleles
gave rise to a child with functional epidermal lies bull nose. The infant has tiresome 1, where he
inherited two recessive alleles from his mother. A translocation between chromosome 1 and
chromosome 22 generated a recessive phenotype. Plasmid DNA, like bacterial DNA: is circular.
is single stranded. comes in homologous pairs. All of the above. None of the above.
Solution
10. Answer is B. EFG of the wild type will form a deletion loop. Because EFG in one
chromosome has no matching pair in other chromosome. So EFG loop will be formed and it will
be removed.
13. Answer is A. Plasmid DNA like bacte.
Directions Problem 1. A female with Muppetrus bristle mates with a .pdf
Directions: Problem 1. A female with Muppetrus bristle mates with a male with Flonasean body
color, producing the following counts: F1 Generation Males Disease-Disease WIT-Disease WT-
WT Disease-WT 243 239 Females WT-Disease Disease Disease WT-WT Disease-WT 254 254
Solution
Parents are heterozygous. Given that each parent has one disease phenotype at one locus and
wild type at other locus.
Female\'s disease phenotype is Muppetrus bristle. Let its symbol be WT.
Cross is: +/disease (phenotype Muppetrus) crossed to +/disease (phenotype Flonasean)
The answer cannot be homozygous lethal because WT-WT and Disease-Disease are there.
Similarly it cannot be autosomal recessive, because in females Disease-Disease is 0.
This means that the mode of inheritance is sex-linked recessive.
In Drosophila males, crossing over does not occur. Only one X chromosome, which comes from
the female parent. So, genotype of males will be \"WT (wild type)\" phenotype on X-
chromosome. In females Disease-Disease is zero because of this..
Problem 13.32 Copy For the following crosses, determine as accurately.pdf
Present value calculations allow one to determine
A. the return to an uncertain asset.
B. the present-day costs and/or benefits of various options.
C. the utility value of a particular option.
D. the social cost of financial calculations.
E. the real wage.
Solution
B. the present-day costs and/or benefits of various options.
Present value calculations involve discounting future costs and/or benefits of a project so that
they can be compared with other available alternatives..
Linked inheritance
Breaking down Biology into simpler bits is the most effective way to learn hence this presentation aims to simplify the concept of 'Linked Inheritance' which makes understanding Inheritance better.
match the following attributes to the parts of a compilerstrips ou.pdf
match the following attributes to the parts of a compiler
strips out the comments and whitespace
converts text into lexemes
generates an Abstrat Symbol Tree
recursive descent or table driven
uses BNF or EBNF definitions
a. lexer
b. parser
Solution
The Program and Portfolio Management Maturity Model is an effective tool for organizations to
decide quickly what PPM improvements they should make to enhance their organization\'s
ability to optimize investments, execute big changes and deliver value.
Overview
Key Findings
Real project, program and portfolio management, when operating at the level of well-integrated
practices, is the key enabler that allows organizations to identify and execute strategic change.
Any meaningful undertakings to enhance or evolve the program and portfolio management
(PPM) function must pay more than lip service to organizational structure, business model and
culture to have any chance of success.
Competitive pressures and changing market conditions are forcing organizations toward Level 3,
and most organizations still aren\'t ready to make the leap.
Recommendations
At any level of PPM maturity, focus first on helping the organization make well-informed
investment choices. This will improve the odds of project success more than any other factor.
For organizations that require enterprisewide change and capabilities, it is worth the effort to
pursue Level 4, where enterprise PPM practices are built.
Identify objectives for your enterprise, and use them to identify the most salient improvement
opportunities for your enterprise.
Meticulously manage the change involved in maturing/improving your PPM capabilities.
Analysis
Project/program/portfolio management office (PMO) leaders, program managers and portfolio
managers are frequently challenged by ineffective processes, lack of stakeholder engagement and
difficult-to-quantify value. 1 Part of the problem is failure to match their processes, people and
technology approaches to the maturity level of their organizations. 2
The Gartner Program and Portfolio Management (PPM) Maturity Model assessment is designed
to help PPM leaders understand best practices around large project management, as well as PPM
to handle delivery of strategic value. This model assumes that organizations progress through a
maturity curve and that each level of organizational maturity directly affects the level of
investment and types of PPM approaches organizations choose to adopt.
arsing
A parser is an algorithm that determines whether a given input string is in a language and, as a
side-effect, usually produces a parse tree for the input. There is a procedure for generating a
parser from a given context-free grammar.
Recursive-Descent Parsing
Recursive-descent parsing is one of the simplest parsing techniques that is used in practice.
Recursive-descent parsers are also called top-down parsers, since they construct the parse tree
top down (rather than bottom up).
The basic idea of recursive-descent par.
Multiple-Choice Which one is a discrete variableRadius of a ball.pdf
Multiple-Choice: Which one is a discrete variable?
Radius of a ball in an urn
Weight of a ball in an urn
Diameter of a ball in an urn
Number of balls in an urn
Solution
A discrete variable can only take up some specific values.
Hence,
OPTION D: Number of balls in an urn [ANSWER]
as it can only take integer values..
More Related Content
Similar to Ferdinand is affected by hemophilia, a recessive x-linked condition..pdf
uetator Periodic Table Question 6 of 30 Incorrect Incorrect DA Saplin.pdf
uetator Periodic Table Question 6 of 30 Incorrect Incorrect DA Sapling Learning sponsible
syndrome is anX-inked, recessive condition caused by mutation of the HPRT1 muscle
movements, The syndrome is characterized byjuvenile-onset gout, in and behavioral disorders. A
mother and father have a young daughter who was diagnosed with Lesch-Nyhan syndrome. No
one in the with family was ever diagnosed the The maternal grandfather was diagnosed
syndrome, but neither one of her parents was diagnosed with condition. Select all the
descriptions that could explain how their daughter inherited Lesch-Nyhan syndrome. The
daughter could not have Nyhan because her maternal grandparents did not have disease. The girl
must have acquired the disease through a random mutation. The daughter inherited two
xchromosomes with the of the mothers X mutation. Nondisjunction chromosome chromosomes
occurred during meiosis, and the daughter inherited from her father. The father did not contribute
a sex chromosome to his daughter due to nondisjunction of the sex chromosomes. The daughter
is xo and her only X chromosome came from her mother, who was a O The d from her father,
who was a carrier. camies the Lesch-Nyhan mutation and was Her inherited There is a hint
available! bottom divider bar. Click on the divider bar again to hide the hint. Previous ® Give
Up & View solution Check Answer 0 Next HExi Hint
Solution
1) the hint here is that the mother\'s maternal was diagnosed with Lesch-Nyhan syndrome. Since
it is an X-chromosome related disease, it is very rare to have females affected by it. If a girl has
LNS, she must have inherited two effected X chromosomes from each parent. Also, one third of
cases arise from new mutations even when there is no family history of any. So the first two
points are the possible reasons for the girls to have LNS..
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..
sex linked inheritance, Sex Influence inheritance and sex limited characters
sex linked inheritance, Sex Influence inheritance and sex limited characters
Sex-linked inheritance by Puzon and Tope
Sex-linked inheritance by Puzon, kathleen Faye R. and Tope, Donita J.
Sex Linkage-Unique Heritage
This presentation intends to explore the sex-linked characters along with some fatal diseases of human beings, their cause, consequences and other issues.
In Drosophila melanogaster, the yellow body gene is X-linked. The yel.pdf
In Drosophila melanogaster, the yellow body gene is X-linked. The yellow body phenotype is
recessive to brown body phenotype. A yellow body female fruit fly was mated to a brown body
male. The vast majority of the male progeny are yellow body. However, a rare brown body male
was found in the progeny (1 out of 2060 males). What is the likely karyotype of this brown body
son? XXY XX XYY XO XXX
Solution
Sex inheritance by modified Mendelian mechanism:
X-linked inheritance means that the gene causing the trait or the disorder is located on the X
chromosome.
X-linked recessive inheritance is a mode of inheritance in which a mutation in a gene on the X
chromosome causes the phenotype to be expressed in males who are necessarily hemizygous for
the gene mutation because they have one X and one Y chromosome. Carrier females who have
only one copy of the mutation do not usually express the phenotype,
Females are XX (homogametic)
Males are XY (heterogametic)
Body color inherited by single gene, two allele system, brown dominant to yellow
But, sex and body color not inherited independently
Body color is linked to sex
Body color gene on X chromosome
Therefore, males have only one body color allele: Haploid and females have two eye color
alleles: Diploid.
Normally all the sons and none of the daughters show the recessive sex-linked characters of the
mother when the father carries the dominant allelomorph. For example, a yellow body female
mated to a brown body male produces yellow body males and brown body females, but rarely
also a yellow body female or a brown body son. The production of these exceptions by a normal
XX female must be due to an aberrant reduction division at which the two X chromosomes fail
to disjoin from each other. In consequence both remain in the egg or both pass into the polar
body. In the latter case an egg without an X chromosome is produced. Such an egg fertilized by
an X sperm produces a male with the constitution XO. These males received their single X from
their father and therefore show the father\'s characters. But that the Y does play some positive
role is proved by the fact that all the XO males have been found to be absolutely sterile.
Answer is d) XO.
A wild-type chromosome has the following segments A B C middot D E F.pdf
A wild-type chromosome has the following segments: A B C middot D E F G H I An individual
is heterozygous for the following chromosome mutation: A B C middot D H I How would the
mutated and wild-type chromosomes pair in prophase I of meiosis? D E F G H of the wild-type
chromosome will form a deletion loop. E F G of the wild-type chromosome will form a deletion
loop. D E F G H of the wild-type chromosome will form an Inversion loop. E F G of the wild-
type chromosome will form an inversion loop. Red-green color blindness is a human x-linked
recessive disorder. A young man with a 47, XXY karyotype (Klinefelter syndrome) is color
Wind. His 46, XY brother also is color blind. Both parents have normal color vision. Where did
the no disjunction that gave rise to the young man with Klinefelter syndrome take place? Assume
that no crossing over took place In prophase I of meiosis. Mother - meiosis I Mother - meiosis II
Father - meiosis I Father - meiosis II Both parents Functional epidermal lies bull nose (JEB) Is a
severe skin disorder that results in blisters over the entire body. The disorder is caused by
autosomal recessive mutations at any one of three loci that help to encode lamina 5, a major
component In the dermal-epidermal basement membrane. Leena Pulicine and colleagues
described a male newborn who was born with JEB and died at 2 months of age (L. Pulkkinen et
al. 1997. American Journal of Human Genetics 61:611-619); the child had healthy unrelated
parents. Chromosome analysis revealed that the infant had 46 normal-appearing chromosomes.
Analysis of DNA showed that his mother was heterozygous for a JEB-causing allele at the
LAMB3 locus, which is on chromosome 1. The father had two normal alleles at this locus. DNA
fingerprinting demonstrated that the male assumed to be the father had, In fact, conceived the
child. Assuming that no new mutations occurred in this family, what might best explain the
presence of an autosomal recessive disease in the child when the mother is heterozygous and the
father Is homozygous normal? The infant inherited both copies of chromosome 1 from the
mother, which is an example of unit parental disomic. The infant is monatomic for chromosome
1, which he received from his mother but did not receive a chromosome 1 from his father.
Epitasis masked the disease phenotype in the father, but complementation of the parental alleles
gave rise to a child with functional epidermal lies bull nose. The infant has tiresome 1, where he
inherited two recessive alleles from his mother. A translocation between chromosome 1 and
chromosome 22 generated a recessive phenotype. Plasmid DNA, like bacterial DNA: is circular.
is single stranded. comes in homologous pairs. All of the above. None of the above.
Solution
10. Answer is B. EFG of the wild type will form a deletion loop. Because EFG in one
chromosome has no matching pair in other chromosome. So EFG loop will be formed and it will
be removed.
13. Answer is A. Plasmid DNA like bacte.
Directions Problem 1. A female with Muppetrus bristle mates with a .pdf
Directions: Problem 1. A female with Muppetrus bristle mates with a male with Flonasean body
color, producing the following counts: F1 Generation Males Disease-Disease WIT-Disease WT-
WT Disease-WT 243 239 Females WT-Disease Disease Disease WT-WT Disease-WT 254 254
Solution
Parents are heterozygous. Given that each parent has one disease phenotype at one locus and
wild type at other locus.
Female\'s disease phenotype is Muppetrus bristle. Let its symbol be WT.
Cross is: +/disease (phenotype Muppetrus) crossed to +/disease (phenotype Flonasean)
The answer cannot be homozygous lethal because WT-WT and Disease-Disease are there.
Similarly it cannot be autosomal recessive, because in females Disease-Disease is 0.
This means that the mode of inheritance is sex-linked recessive.
In Drosophila males, crossing over does not occur. Only one X chromosome, which comes from
the female parent. So, genotype of males will be \"WT (wild type)\" phenotype on X-
chromosome. In females Disease-Disease is zero because of this..
Problem 13.32 Copy For the following crosses, determine as accurately.pdf
Present value calculations allow one to determine
A. the return to an uncertain asset.
B. the present-day costs and/or benefits of various options.
C. the utility value of a particular option.
D. the social cost of financial calculations.
E. the real wage.
Solution
B. the present-day costs and/or benefits of various options.
Present value calculations involve discounting future costs and/or benefits of a project so that
they can be compared with other available alternatives..
Linked inheritance
Breaking down Biology into simpler bits is the most effective way to learn hence this presentation aims to simplify the concept of 'Linked Inheritance' which makes understanding Inheritance better.
Similar to Ferdinand is affected by hemophilia, a recessive x-linked condition..pdf (20)
uetator Periodic Table Question 6 of 30 Incorrect Incorrect DA Saplin.pdf
uetator Periodic Table Question 6 of 30 Incorrect Incorrect DA Saplin.pdf
With X-linked inheritance, the X chromosome is transmitted in a diffe.pdf
With X-linked inheritance, the X chromosome is transmitted in a diffe.pdf
sex linked inheritance, Sex Influence inheritance and sex limited characters
sex linked inheritance, Sex Influence inheritance and sex limited characters
In Drosophila melanogaster, the yellow body gene is X-linked. The yel.pdf
In Drosophila melanogaster, the yellow body gene is X-linked. The yel.pdf
A wild-type chromosome has the following segments A B C middot D E F.pdf
A wild-type chromosome has the following segments A B C middot D E F.pdf
Directions Problem 1. A female with Muppetrus bristle mates with a .pdf
Directions Problem 1. A female with Muppetrus bristle mates with a .pdf
Problem 13.32 Copy For the following crosses, determine as accurately.pdf
Problem 13.32 Copy For the following crosses, determine as accurately.pdf
More from arpitaeron555
match the following attributes to the parts of a compilerstrips ou.pdf
match the following attributes to the parts of a compiler
strips out the comments and whitespace
converts text into lexemes
generates an Abstrat Symbol Tree
recursive descent or table driven
uses BNF or EBNF definitions
a. lexer
b. parser
Solution
The Program and Portfolio Management Maturity Model is an effective tool for organizations to
decide quickly what PPM improvements they should make to enhance their organization\'s
ability to optimize investments, execute big changes and deliver value.
Overview
Key Findings
Real project, program and portfolio management, when operating at the level of well-integrated
practices, is the key enabler that allows organizations to identify and execute strategic change.
Any meaningful undertakings to enhance or evolve the program and portfolio management
(PPM) function must pay more than lip service to organizational structure, business model and
culture to have any chance of success.
Competitive pressures and changing market conditions are forcing organizations toward Level 3,
and most organizations still aren\'t ready to make the leap.
Recommendations
At any level of PPM maturity, focus first on helping the organization make well-informed
investment choices. This will improve the odds of project success more than any other factor.
For organizations that require enterprisewide change and capabilities, it is worth the effort to
pursue Level 4, where enterprise PPM practices are built.
Identify objectives for your enterprise, and use them to identify the most salient improvement
opportunities for your enterprise.
Meticulously manage the change involved in maturing/improving your PPM capabilities.
Analysis
Project/program/portfolio management office (PMO) leaders, program managers and portfolio
managers are frequently challenged by ineffective processes, lack of stakeholder engagement and
difficult-to-quantify value. 1 Part of the problem is failure to match their processes, people and
technology approaches to the maturity level of their organizations. 2
The Gartner Program and Portfolio Management (PPM) Maturity Model assessment is designed
to help PPM leaders understand best practices around large project management, as well as PPM
to handle delivery of strategic value. This model assumes that organizations progress through a
maturity curve and that each level of organizational maturity directly affects the level of
investment and types of PPM approaches organizations choose to adopt.
arsing
A parser is an algorithm that determines whether a given input string is in a language and, as a
side-effect, usually produces a parse tree for the input. There is a procedure for generating a
parser from a given context-free grammar.
Recursive-Descent Parsing
Recursive-descent parsing is one of the simplest parsing techniques that is used in practice.
Recursive-descent parsers are also called top-down parsers, since they construct the parse tree
top down (rather than bottom up).
The basic idea of recursive-descent par.
Multiple-Choice Which one is a discrete variableRadius of a ball.pdf
Multiple-Choice: Which one is a discrete variable?
Radius of a ball in an urn
Weight of a ball in an urn
Diameter of a ball in an urn
Number of balls in an urn
Solution
A discrete variable can only take up some specific values.
Hence,
OPTION D: Number of balls in an urn [ANSWER]
as it can only take integer values..
Looking for some help with this assignment. 1. Define distribut.pdf
Looking for some help with this assignment.
1. Define distributed computing and the impact it has had on organizations.
2. Why should you be concerned about the innovation butterfly is you are an IT professional?
3. What is the difference between single threading and multithreading and how would I use it in
a Distributed system?
4. What is a mobile architecture and what are some of the design considerations?
5. Explain why naming conventions are important in large enterprise organizations.
6. What is the difference between a disruptive and a sustaining technology?
7. Describe the difference amongst Mobile, Cloud and Embedded systems.
8. Explain the role time plays in distributed systems and why it is difficult to manage.
9. What are the major challenges IOT plays in a distributed system?
10. How does mobility, ubiquity and PAN increase the complexity of distributed systems.
Solution
1)
sol:defnition:
it is a one type of model where software components are shared with mutliple computer
to make work easier and gives more effieciency and performance
impact of distributed computing
1)at present it keeps a main roles in organizations ,all the organizations are converting to
distributed system
2)we can process data at one place and access data at one place so there is lot of availability
3)many organization doing thier business process in remote computer and accesing their
database
2)differenece between sustaining and disruptive technology
1)by using sustaining technology improves performance for products which had been established
with the
dimension of the product which gives market to the mainstream customers
2) coming to the disruptive technology bring to market in diffrent praportions of values which
had been
available previously,these disruptive techonolgy underperform which had been established in
main steram market.
Introduction and BackgroundIn recent lectures we discussed usi.pdf
Introduction and Background
In recent lectures we discussed using arrays, classes and interfaces (see newly added course
notes if you want to read ahead about interfaces – we will cover them this week in lecture). In
this lab you will utilize all of these topics to build a simple yet useful new class. Consider the
following interface describing the methods for a simple double ended queue (or deque):
public interface SimpleDeque
{
public void addFront(Object X); // Add Object X at front of list
public void addRear(Object X); // Add Object X at rear of list
// If array is full, add methods should do nothing
public Object removeFront(); // Remove and return Object X from
// front of list
public Object removeRear(); // Remove and return Object X from
// rear of list
// If array is empty, remove methods should return null
public boolean isEmpty(); // Return true if the list is empty
// Return false otherwise
}
A queue has the behavior such that items are added at the rear and removed from the front,
thereby giving a First In First Out (FIFO) access to the items added and subsequently removed
from the list. No other manipulations of the data are permitted (for example, we cannot add or
remove anywhere in the middle). Looking at it \"in reverse\", we could add new items at the
front of the queue and remove them from the rear. This is still providing FIFO access, but just
from a different point of view. Now consider both adding and removing items at the rear of the
list (without ever accessing the front). This is called stack access and gives us Last In First Out
(LIFO) access to the items (the data is removed in reverse order). The same behavior occurs if
we both add and remove at the front without ever accessing the rear of the list.
The simple deque above is expressed as an interface rather than a class, because we are not
describing the data or how it is represented -- we are simply describing its access behavior.
However, to actually build a working deque, we need a class that implements the interface
above. For example:
public class MyDeque implements SimpleDeque
{
Object [] theData;
int numItems;
public MyDeque(int maxItems)
{
theData = new Object[maxItems];
numItems = 0;
}
// Implementation of the five methods of SimpleDeque, plus
// perhaps other methods as well
}
Note that the implementation above uses an array of Object to store the items in the deque. Since
Object is the base class to all other Java classes, an array of Object can thus be used to store any
Java class types (we can even store primitive values if we utilize their wrapper classes). Also
note that nothing in the SimpleDeque interface requires an array to be used to store the data. You
will see in your CS 0445 course that a linked list may in fact be a better implementation than an
array in this case. However, for this implementation we will use an array because it is simple and
easy to understand.
Another important thing to notice about the partial implementation above is tha.
In the binary search, if the array being searched has 32 elements in.pdf
In the binary search, if the array being searched has 32 elements in it, how many elements of the
array must be examined to be certain that the array does not contain the key? What about 1024
elements? Note: the answer is the same regardless of whether the algorithm is recursive or
iterative.
Solution
Binary Search Algorithm- Fundamentals, Implementation and Analysis
Hitesh Garg | May 15, 2015 | algorithms | 5 Comments
Binary Search Algorithm and its Implementation
In our previous tutorial we discussed about Linear search algorithm which is the most basic
algorithm of searching which has some disadvantages in terms of time complexity,so to
overcome them to a level an algorithm based on dichotomic (i.e. selection between two distinct
alternatives) divide and conquer technique is used i.e. Binarysearch algorithm and it is used to
find an element in a sorted array (yes, it is a prerequisite for this algorithm and a limitation too).
In this algorithm we use the sorted array so as to reduce the time complexity to O(log n). In this,
size of the elements reduce to half after each iteration and this is achieved by comparing the
middle element with the key and if they are unequal then we choose the first or second half,
whichever is expected to hold the key (if available) based on the comparison i.e. if array is sorted
in an increasing manner and the key is smaller than middle element than definitely if key exists,
it will be in the first half, we chose it and repeat same operation again and again until key is
found or no more elements are left in the array.
Recursive Pseudocode:
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// initially called with low = 0, high = N – 1
BinarySearch_Right(A[0..N-1], value, low, high) {
// invariants: value >= A[i] for all i < low
value < A[i] for all i > high
if (high < low)
return low
mid = low +((high – low) / 2) // THIS IS AN IMPORTANT STEP TO AVOID BUGS
if (A[mid] > value)
return BinarySearch_Right(A, value, low, mid-1)
else
return BinarySearch_Right(A, value, mid+1, high)
}
Iterative Pseudocode:
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BinarySearch_Right(A[0..N-1], value) {
low = 0
high = N - 1
while (low <= high) {
// invariants: value >= A[i] for all i < low
value < A[i] for all i > high
mid = low +((high – low) / 2) // THIS IS AN IMPORTANT STEP TO AVOID BUGS
if (A[mid] > value)
high = mid - 1
else
low = mid + 1
}
return low
}
Asymptotic Analysis
Since this algorithm halves the no of elements to be checked after every iteration it will take
logarithmic time to find any element i.e. O(log n) (where n is number of elements in the list) and
its expected cost is also proportional to log n provided that searching and comparing cost of all
the elements is same
Data structure used -> Array
Worst case performance -> O(log n)
Best case performance -> O(1)
Average case performance -> O(log n)
Worst case space complexity -> O(1)
So the idea is-
RECURSIVE Implementation of Binary search in C programming language
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Is there any information out there about decorator crabs picking up .pdf
Is there any information out there about decorator crabs picking up toxins and the correlation to
the polluted areas they inhabit
Solution
It\'s not true that these crabs pick up toxins from the environment. They use the surviving
environment to camouflage themselves. The is also no relation to the pollution thing..
Heliconius melpomene is a species of butterfly with ZW sex determina.pdf
Heliconius melpomene is a species of butterfly with ZW sex determination. H. melpomene\'s
wing spots are normally orange, but a recessive Z-linked mutation causes the spots to be yellow,
while a recessive autosomal mutation causes the spots to be red. Butterflies homozygous (or
hemizygous) for both mutations have no wing spots. What are the expected F1 and F2
phenotypes from a cross between a red-spotted female and a yellow-spotted male?
Solution
The Heliconius melpomene butterfly having sex determination ZW have normally orange wing
spots.
Given
Recessive Z linked mutation causes the spots to be yellow, if Y is the dominant gene the y
represents the recessive gene. The recessive autosomal mutation causes the spot to be red,the
recessive autosomal mutation represented by rr.
a) WX represents female
b) WW represents male
Given-:
F1 generation
The red spotted female is crossed with yellow spotted male
ZWrr X ZZy
ZZ, WrrZ, ZZy, WrrZy
It gives 1 hemizygous butterfly, 1red spotted female butterfly, 1 yellow spotted male and 1
normal orange wing male butterfly.
The F2 generation gives the same result..
Identify the key mechanism that determines which X remains active an.pdf
Identify the key mechanism that determines which X remains active and which X becomes
inactive.
Solution
In normal human females, only one X chromosome is transcriptionally active, the other X
chromosome remains condensed as heterochromatin and is called as sex chromatin or Barr body.
Formation of Barr body ensures that cells of both males and females have same number of active
X chromosomes and thus synthesise equal amount of products encoded by X linked genes.
This inactivation of one of the two X chromosome in each cell is called Lyonization, after the
name of British researcher Mary Lyon. She proposed that;
Inactivation of X chromosome in females occurs during early stages of embryonic development
and this Inactivation of X chromosome in embryo is a random process, so that paternally derived
and maternally derived X chromosomes stands equal chance of becoming inactivated in any
given cell.
MECHANISM OF INACTIVATION :
The mechanism of Inactivation is initiated by a non coding RNA molecule called XIST RNA.
There is a region on X chromosome called as X-Inactivation center (XIC) located on proximal
end of X chromosome. Genetic expression of this region occurs only on chromosome that is
inactivated. This region carries about 4 genes one of which is X-Inactivation specific transcript
(XIST). The RNA transcribed by this gene is called XIST RNA. This RNA is large (about 17kb
long) and is not translated. It does not diffuse into nucleoplasm,but accumulates along the length
of chromosome just before it is Inactivated.
Thus RNA product of XIST gene spread over and coat the X chromosome bearing the gene that
produced it ,creating some sort of molecular \"cage\" that entraps it leading to its Inactivation.
Thus X chromosome in which XIST gene is transcribed to form XIST RNA becomes inactive
(due to coating of this X chromsome by XIST RNA) while as the other X chromosome remains
active..
For each of the given functions, list the domain and range. Also list.pdf
For each of the given functions, list the domain and range. Also list the period, amplitude and
the equations of all asymptotes where applicable. (If you choose to draw a graph it will not be
graded.) ln(x) Domain:_ Range:_ Period:_ Amplitude: _ Asymptotes:_
Solution
f(x) = lnx
Domain : ( 0, infinity)
Range : ( -inf , +inf)
Period : It is not periodic function
Amplitude : Vertical asymtote , x=0.
Given to girls born in the U.S. in the years The tables below give t.pdf
Given to girls born in the U.S. in the years The tables below give the ten most popular names
1974 and 1999 and the percentages of girls given those names. Consider group of 20 girls born in
1974. and another group of 20 bom in 1999. Use the data in the tables to do problems 22-25
regarding these groups. The name Jennifer was approximately 3 times as popular in 1974 as
Emily was in 1999, despite each being the #1 name in their respective years. Determine by
calculation if you are 3 times as likely to have more than one Jennifer in the 1974 group of 20 as
you are to have multiple Emilys in the 1999 group of 20.
Solution.
How do open source projects work How is it different from closed sou.pdf
How do open source projects work? How is it different from closed source projects? Discus the
creation of a database. Explain its working. (E.g, what you need to use to create tables and
relationship? What you do to modify a database?) Explain what\'s SaaS, PaaS and laaS?
Distinguish among these three.
Solution
31) In open source project the source code of the project is made available to anyone for free.
The code can be studied, modified and even redistributed for any purpose.Many of the open
source projects are developed in an open collaberative manner. People volunteer to develop the
projects.Where as in closed source projects the source code is not developed by an organization
and the source code is not openly available.
32) A database consists of many tables created for a project. In mysql it is created using the
command \'create database sample_database;\'. We generally use a RDBMS(Relational Database
Management System). The data is stored in RDBMS database objects called tables. You can
create a table by specifying the number of rows, names of each row and the type of data that will
be stored in each of them.
33) SaaS means software as a service in which a software is provided as a serive. Generally the
service is centrally hosted, but this many not always be the case. The consumer is charged on a
one time basis or in a subscription model. PaaS stands for platform as a service. The service
provides platform as a service for customers to develop, manage and run applications. This is
generally referred for cloud computing applications. LaaS stands for logging as a serivce. This
represents a architechture used for collection of log files from a specific location or source..
Explain whether you think IFRS and GAAP will be fully converged. Wha.pdf
Explain whether you think IFRS and GAAP will be fully converged. What is the current status of
convergence projects?
Solution
There has been a concerted convergence effort between the Generally Accepted Accounting
Principles (GAAP) and the International Financial Reporting Standards (IFRS) in order to avoid
conflict and confusion, promote simplicity, streamlining, consistency and transparency, and
avoid any future financial crises or meltdowns.
Despite the research-indicated evidence of a higher accounting quality being experienced by
firms that either apply the IFRS standards or have switched to them from the GAAP, the
convergence process has not proven to be an easy task, mostly because of the differences in
approach between the two accounting bodies.
The GAAP is a rules-based methodology, while the IFRS takes a principle-based approach. The
rules-based approach is comprised of a complex set of guidelines that establishes criteria for
every possible contingency and provides the rules required for specified transactions, thus
promoting uniformity. The principle-based methodology lays out the key objectives of good
reporting in each subject area and then provides guidance, explaining the objective, and relates it
to common examples, thus promoting transparency.
If these methodological differences between the two approaches cannot be resolved, they may
prolong the process of compiling a true set of international accounting standards and increase the
costs required to maintain two sets of books.
So from the above discussion it is clear that, IFRS and GAAP will be fully converged, but it will
took a time.
current status of convergence projects
One of the main concerns in the United States business world is how the convergence process
and its results will impact the future evolution of the accounting profession. This specific
concern, simply stated, is about uniformity over transparency, and it has a serious impact on the
standards development process. Could the goals of uniformity and transparency be achieved?
Are they incompatible or mutually exclusive?
This incompatibility - real or perceived - is grounded in the conflicts existing among the
constructs of rules-based and principle-based shareholder and stakeholder primacy theories,
which are recognized by the Financial Accounting Standards Board (FASB), the International
Accounting Standards Board (IASB) and the European and Asian Accounting Standards Boards,
and which have an impact on the standards development methodology. Transparency has a direct
impact on the areas of business combinations (Phase I and II), revenue recognition and financial
performance of business enterprises reporting. hence they are still in process..
Does the following picture depict a function It so, is it a one to o.pdf
Does the following picture depict a function? It so, is it a one to one function?
Solution
The given picture depict the function.Every element of set x mapping into set y\'s exactly one
element.so it is called a function.
A function is said to be one-to-one if every y value has exactly one x value mapped onto it.So the
given picture is not one-to-one function.because two elements of set x mapping to one set y\'s
element..
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Do patients have health care rights that are not legal or statuatory rights? Explain your answer.
Solution
Legal Conceptualization: it supplies that a patient has a right to realize potential benefits, risks,
and option a of proposed therapy; duty of physician to reveal to the patient a full apprehension of
the suggested care before patient assent to it.
Statuatory Rights: when a patient is clinically unfit to give assent to a lifesaving emergency
therapy. The law implicit assent (exigency in most states abolish need for assent).
Describe the authentication process in the figure below Solutio.pdf
Describe the authentication process in the figure below:
Solution
Authentication begins when a user tries to access information. First, the user must prove his
access rights and identity. When logging into a computer, users commonly enter usernames and
passwords for authentication purposes. This login combination, which must be assigned to each
user, authenticates access. However, this type of authentication can be circumvented by hackers.
A better form of authentication, biometrics, depends on the user’s presence and biological
makeup (i.e., retina or fingerprints). This technology makes it more difficult for hackers to break
into computer systems.
The Public Key Infrastructure (PKI) authentication method uses digital certificates to prove a
user’s identity. There are other authentication tools, too, such as key cards and USB tokens. One
of the greatest authentication threats occurs with email, where authenticity is often difficult to
verify. For example, unsecured emails often appear legitimate..
Discuss the history of bureaucratic and administrative management..pdf
Discuss the history of bureaucratic and administrative management.
Solution
1. History of Bureaucratic Management:
There was time before 1800 when the organisations operated like a family. Supervisors, workers
and employees did not have proper defined work. Every one did everything. During late 1800,
Max Weber brought about changes in this system of organisational operation. He believed that
having an informal organisation decreases the productivity of the employees as they are not
focussed and goal-driven.
Max Weber created the idea of bureaucratic management which made the organisation more
rigid, authorative and structured. He believed that employees do not follow the organisation but
they follow their bosses. He defined supervisors and workers role properly. The workers were
supposed to take their immediate bosses orders and cannot operate beyond that capacity.
Max Weber\'s theory of bureaucratic management has two important elements. First, it involves
structuring an organization into a hierarchy system. Secondly, the organization as well as its
employees operate through clearly defined rational-legal decision-making rules. Both the
elements help the organisation to achieve its goals.
2. History of Administrative Mangement
During 1900, the management expert Jules Henri Fayol, came to the conclusion that
management is an activity that is followed by all the human undertakings. Management is
needed at school, housework, business, government services, etc. All these human undertakings
need five basic administrative functions to operate. They are:
i. Planning
ii. Organizing
iii. Commanding
iv. Coordinating
v. Controlling
Henri Fayol rejected the notion that \"Managers are born and not made.\" Instead he believed that
management is a skill that can be developed over the time through proper training and
enthusiasm. Managemnt skills can be acquired if its principles are understood.
Henri Fayol was of the belief that since management is an all-encompassing activity, the
management skills should be taught in schools, colleges and universities..
Describe or explain how SNPs in genes at evolutionary conserved brea.pdf
Describe or explain how SNPs in genes at evolutionary conserved breakpoints can lead to
dramatic phenotypic differences between species like chimpanzees and humans even though
their genomes are 98% similar? Provide and describe or explain an example of one gene with
SNP differences between gorillas and humans that might have resulted changes between humans
and apes.
Solution
SNP ( Single nucleotide polymorphism) is a type of genetic variation among living beings. SNP
represents a difference in a single DNA building block i.e, nucleotide. SNP replaces the
nucleotide cytosine ( C) With ghe nucleotide thymine (T) in a stetch of DNA. The difference
between the two genomes of humans and chimpanzees is actually 4% approximately. This
msinly contributes to the phenotypic differences in various species.
Humans differbfrom other primates like chimpanzees and apes in cognitive abilities and a
significantly larger brain. These correlate with metabolic changes, etc. Example of such genes
with SNP differences is 90 TF genes with significant expression levels in humans and
chimpanzee brain. Among these KRAB-Zinc finger genes are markedly important. The
differentially expressed TFs cluster within robust regulatory network consisting of two distinct
but interlinked modules, strongly associated with energy metabolism functions and the others
with transcription, vesticular transport and ubiquitination..
Data x word 23 y word -3 Write assembly code for the following p.pdf
Data x: word 23 y: word -3 Write assembly code for the following pseudo code and draw flow
chard (for both Q1 and Q2): if (x
Solution
Answer:
Assembly Language Code :
1)
.zero 1
main:
push rbp
mov rbp, rsp
cmp DWORD PTR [rbp-4], 4
jg .L2
add DWORD PTR [rbp-8], 1
2)
.zero 1
main:
push rbp
mov rbp, rsp
mov DWORD PTR [rbp-4], 1
mov eax, 1
test al, al
je .L2
mov DWORD PTR [rbp-8], 0.
Can Variance Inflation factors be used to detect Heteroscedasticity .pdf
Can Variance Inflation factors be used to detect Heteroscedasticity problem in the regression
analysis
Solution
Variance inflation factors (VIF) measure how much the variance of the estimated regression
coefficients are inflated as compared to when the predictor variables are not linearly related.
Use to describe how much multicollinearity (correlation between predictors) exists in a
regression analysis. Multicollinearity is problematic because it can increase the variance of the
regression coefficients, making them unstable and difficult to interpret.
Use the following guidelines to interpret the VIF:VIFStatus of predictorsVIF = 1Not correlated1
< VIF < 5Moderately correlatedVIF > 5 to 10Highly correlated.
A study undertaken by the Miami-Dade Supervisor of Elections in 2002.pdf
A study undertaken by the Miami-Dade Supervisor of Elections in 2002 revealed that 44% of
registered voters are Democrats, 37% are Republicans, and 19% are others. If two registered
voters are selected at random, find the following probability:
One of them is a Democrat and another one is a Republican.
Solution
Here we have following probabilties:
P(democrats)=0.44
P(republicans)=0.37
P(others)=0.19
Since voters are independent from each other so the probabilityOne of them is a Democrat and
another one is a Republican is
P(One of them is a Democrat and another one is a
Republican)=P(democrats)P(republican)=0.44*0.37=0.1628
hence, the required probability is 0.1628..
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