A mother and father both have normal karyotypes. The father is color .pdf
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A mother and father both have normal karyotypes. The father is color blind, but the mother is not. Color blindness is determined by a recessive mutation on the X chromosome. Their first child is a color-blind girl with an XX karyotype. The family consults a genetic counselor because their second child is a color blind boy with Klinefelter\'s syndrome. Klinefelter\'s syndrome results when a child has the sex chromosome constitution XXY. Which of the following statements is true about the nondisjunction event that led to the boy with Klinefelter\'s syndrome (assume no recombination)? Only during the first meiotic division in Dad. Only during the second meiotic division in Dad. Only during the first meiotic division in Mom. Only during the second meiotic division in Mom. Either a or d. Either a orb. Solution ans-e Non-disjunction (failure of homologous chromosomes) at paternal meiosis I produces XY secondary spermatocytes and a 24,XY gamete. Fertilization with a 23,X ovum produce a 47,XXY individual (Klinefelter syndrome). Only nondisjunction at paternal meiosis II produces a 24,YY gamete that yields a 47,XYY individual after fertilization..
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A boy with Klinefelter Syndrome (XXY) has colorblindness, which is a.pdf
A boy with Klinefelter Syndrome (XXY) has colorblindness, which is an X-linked recessive
trait. His mother and father both have normal vision.
Did nondisjunction occur during the formation of the egg or the sperm in this situation? How do
you know?
Solution
Color blindness is an X-recessive trait, there is no way that the father can have color-blindness
gene (let’s say Xc). Hence, the color blindness gene should come from mother (but only a single
copy of gene is present as it is not expressed).
Genotype
Mother: XXc
Father: XY
Child: XcXcY (for trait to be expressed, color blindness gene should be present on both the
copies of X)
Xc can be inherited only from mother (as father doesn’t have it) and both the copies of Xc should
come from her. The egg must have arisen from a nondisjunction in meiosis II. In meiosis I, the
homologous X chromosomes separate, so one cell has X and the other has Xc. Failure of
chromatids to separate in meiosis II would then result in an egg with two copies of Xc.
A man with Klinefelter syndrome (XXY) is found to be colour-blind (a.pdf
A man with Klinefelter syndrome (XXY) is found to be colour-blind (an X-linked recessive
phenotype). Both his parents had normal vision.
a) Explain the stimultaneous origin of Klinefelter syndrome and color blindness by the abnormal
behaviour of chromosome at meiosis.
b) Can your explanation distinguish whether the abnormal chromosomal behaviour occured at
the first or second division of meiosis?
Solution
The question it self is a blender. Because, if the parents have normal vision, In of XXY, the Y
chromosome came from father. The two X chromosomes might came from mother. Here the
Mother had normal vision means she might act as a carrier (Xx). The the same is present in the a
man with Klinefelter syndrome the genotype would me XxY. Due to presence of Normal X
chromosome surely he should have a normal vision.
There is an another chance, that the X chromosome might came form father along with Y
chromosome, at this chance the father should have color blindness. But here the father had
normal vision. So Check your question once and get me back..
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.
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..
frank has klinefelter syndrome (47, XXY)... Frank has Klinefelter sy.pdf
frank has klinefelter syndrome (47, XXY)... Frank has Klinefelter syndrome (47, XXY). His
mother has normal skin, but his father has anhidrotic ectodermal dysplasia, an X-inked condition
where the skin does not contain sweat glands. Frank has patches of normal skin and patches of
skin without sweat glands. Complete the sentences. Some answers may be used more than once
or not at all. Frank received the mutant chromosome from his. Nondisjunction occurred in his
during the meiotic division.
Solution
Answer
Frank recieved the mutant chromosome from his Father
Nondisjunction occured in his Father during the First meiotic division.
Since, the mutant gene is present on X-chromosome of Father of Frank
Frank has also got this disease, therefore, the second chromosome of XXY combination must
have been inherited from father.
Nondisjunction has occured at Meiosis- 1 Stage,
At meiosis -1 chromosomes separate (Not chromatids), therefore, one cell has recieved both
chromosomes (XY)
Each chromosomes is made up of 2 chromatids.
At Meiosis -II chromatids separate. therefore, the sperm from such a cell recieves one chromatids
from X and another chromatids from Y and forms XY combination
when this sperm with XY fuses with egg having X, there resultant zygote has XXY genotype.
I need help working out this question . woman has normal vision al.pdf
I need help working out this question .
woman has normal vision although her maternal grandfather (her mother’s father) had red-
green colorblindness, a sex-linked recessive trait. Her maternal grandmother and the woman’s
own father are assumed to not possess a copy of the mutant allele. The woman marries a man
with normal vision although his father was colorblind. What is the probability that the first child
of this couple will be colorblind?
Solution
As it is x linked inheritance, the genotype of women is either xx = Normal or xxc =Carrier for
disease. If she is carrier (xxc) and married normal man, then the probability of 1st chaild getting
colour blindness is 1/4 = 0.25
Color blind ness is x linked and it inherit criss cross manner. That mean, male transfer color
blindness to her female children and. female transfer her male children.
XXc x XY ----------------> XX, XY, XXc & XcY
XcY progeny is only colour blind. So probability is 1\\4.
Recommended
A boy with Klinefelter Syndrome (XXY) has colorblindness, which is a.pdf
A boy with Klinefelter Syndrome (XXY) has colorblindness, which is an X-linked recessive
trait. His mother and father both have normal vision.
Did nondisjunction occur during the formation of the egg or the sperm in this situation? How do
you know?
Solution
Color blindness is an X-recessive trait, there is no way that the father can have color-blindness
gene (let’s say Xc). Hence, the color blindness gene should come from mother (but only a single
copy of gene is present as it is not expressed).
Genotype
Mother: XXc
Father: XY
Child: XcXcY (for trait to be expressed, color blindness gene should be present on both the
copies of X)
Xc can be inherited only from mother (as father doesn’t have it) and both the copies of Xc should
come from her. The egg must have arisen from a nondisjunction in meiosis II. In meiosis I, the
homologous X chromosomes separate, so one cell has X and the other has Xc. Failure of
chromatids to separate in meiosis II would then result in an egg with two copies of Xc.
A man with Klinefelter syndrome (XXY) is found to be colour-blind (a.pdf
A man with Klinefelter syndrome (XXY) is found to be colour-blind (an X-linked recessive
phenotype). Both his parents had normal vision.
a) Explain the stimultaneous origin of Klinefelter syndrome and color blindness by the abnormal
behaviour of chromosome at meiosis.
b) Can your explanation distinguish whether the abnormal chromosomal behaviour occured at
the first or second division of meiosis?
Solution
The question it self is a blender. Because, if the parents have normal vision, In of XXY, the Y
chromosome came from father. The two X chromosomes might came from mother. Here the
Mother had normal vision means she might act as a carrier (Xx). The the same is present in the a
man with Klinefelter syndrome the genotype would me XxY. Due to presence of Normal X
chromosome surely he should have a normal vision.
There is an another chance, that the X chromosome might came form father along with Y
chromosome, at this chance the father should have color blindness. But here the father had
normal vision. So Check your question once and get me back..
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.
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..
frank has klinefelter syndrome (47, XXY)... Frank has Klinefelter sy.pdf
frank has klinefelter syndrome (47, XXY)... Frank has Klinefelter syndrome (47, XXY). His
mother has normal skin, but his father has anhidrotic ectodermal dysplasia, an X-inked condition
where the skin does not contain sweat glands. Frank has patches of normal skin and patches of
skin without sweat glands. Complete the sentences. Some answers may be used more than once
or not at all. Frank received the mutant chromosome from his. Nondisjunction occurred in his
during the meiotic division.
Solution
Answer
Frank recieved the mutant chromosome from his Father
Nondisjunction occured in his Father during the First meiotic division.
Since, the mutant gene is present on X-chromosome of Father of Frank
Frank has also got this disease, therefore, the second chromosome of XXY combination must
have been inherited from father.
Nondisjunction has occured at Meiosis- 1 Stage,
At meiosis -1 chromosomes separate (Not chromatids), therefore, one cell has recieved both
chromosomes (XY)
Each chromosomes is made up of 2 chromatids.
At Meiosis -II chromatids separate. therefore, the sperm from such a cell recieves one chromatids
from X and another chromatids from Y and forms XY combination
when this sperm with XY fuses with egg having X, there resultant zygote has XXY genotype.
I need help working out this question . woman has normal vision al.pdf
I need help working out this question .
woman has normal vision although her maternal grandfather (her mother’s father) had red-
green colorblindness, a sex-linked recessive trait. Her maternal grandmother and the woman’s
own father are assumed to not possess a copy of the mutant allele. The woman marries a man
with normal vision although his father was colorblind. What is the probability that the first child
of this couple will be colorblind?
Solution
As it is x linked inheritance, the genotype of women is either xx = Normal or xxc =Carrier for
disease. If she is carrier (xxc) and married normal man, then the probability of 1st chaild getting
colour blindness is 1/4 = 0.25
Color blind ness is x linked and it inherit criss cross manner. That mean, male transfer color
blindness to her female children and. female transfer her male children.
XXc x XY ----------------> XX, XY, XXc & XcY
XcY progeny is only colour blind. So probability is 1\\4.
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.
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..
Gregor Mendel used pea plants to study heredity in a series of exper.docx
Gregor Mendel used pea plants to study heredity in a series of experiments. Mendel worked by carefully observing and recording traits in successive generations of plants. Knowledge about DNA and chromosomes came later.
This lab will apply genetic laws to human inheritance using Punnett squares.
Recall that DNA is wound tightly into chromosomes. Cells with only one set of chromosomes, such as sex cells, are
haploid
. When two haploid cells fuse during fertilization, a diploid zygote with two full sets of chromosomes is formed. Most cells of a mature individual are diploid.
Homologous chromosomes
have the same genes, but they might have different versions (
alleles
) of those genes.
Diploid
cells have two alleles for each gene. These alleles might be identical (gene A) or different (gene B). Each gene’s
locus
is its location on a chromosome.
Human traits come through dominant or recessive inheritance. For example, the cystic fibrosis traits carried by a dominant allele are always expressed, even if the recessive gene is present (FF or Ff). The recessive is only expressed when two copies of the recessive allele are present (ff).
Mother: Healthy carrier
F
f
Father: Healthy carrier
F
FF
Healthy non-carrier
Ff
Healthy carrier
f
Ff
Healthy carrier
FF
Affected
Human gender is carried on the X and Y chromosomes. Females are XX and males are XY. Heredity traits such as color blindness, which is the inability to distinguish among some colors, are carried on the X chromosome (X
c
). The presence of one normal X
C
will allow normal vision.
In this next set of exercises, you will determine the genotypes of the parents by considering the inheritance patterns of traits in their children. The following is a table of the phenotypes of the family members:
Phenotype
Alleles
Parents
Mother
Not color blind
Freckles
Type B blood
X
c
X
C
Ff
I
B
i
Father
Color blind
No freckles
Type A blood
X
c
Y
Ff
I
A
i
Children
Abby
Color blind
Freckles
X
c
X
c
Ff or ff
Brady
Not color blind
No freckles
X
C
Y
ff
Carly
Not color blind
No freckles
X
c
X
C
ff
Dennis
Color blind
Freckles
X
c
Y
Ff or ff
Exercise 1: Color Blindness
Using the alleles XC (not color blind) and Xc (color blind), distribute the gametes from each parent to the outside of the Punnett square. Drag and drop the child with the correct phenotype to the box within the Punnett square that has the corresponding genotype that would occur from the fusion of egg and sperm as indicated by your placement of the gametes.
Exercise 2: Freckles
Freckles are groups of cells on the skin that produce the pigment melanin, often in response to exposure to ultraviolet (UV) light. The gene for freckles is inherited in a dominant/recessive pattern. A person carrying even a single copy of the dominant allele (F) will have freckles. A person who is homozygous recessive (ff) will have no freckles.
Using the alleles F (freckles) and f (no freckles), distribute the gametes from each parent to the outside of the Punne.
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.
JAVA ProgrammingDesign a class Message that models an e-mail messa.pdf
JAVA Programming
Design a class Message that models an e-mail message. A message has a recipient, a sender, and
a message text. Support the following methods:
a. A constructor that takes the sender and recipient
b. A method append that appends a line of text to the message body
c. A method toString that makes the message into one long string like this: “From: Fname
Lname\ To: Some Recipient\ …”
Write a driver program that uses this class to make a message and print it.
Solution
MessageTest.java
import java.util.Scanner;
public class MessageTest {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter Recipient Name: \");
String recipientName = scan.nextLine();
System.out.println(\"Enter Sender Name: \");
String senderName = scan.nextLine();
System.out.println(\"Enter the body text: \");
String text = scan.nextLine();
Message m = new Message(recipientName, senderName);
m.append(text);
System.out.println(m.toString());
}
}
Message.java
public class Message {
private String recipient;
private String sender;
private String text;
public Message(String r, String s){
this.recipient = r;
this.sender = s;
text = \"\";
}
public void append(String t){
text = text + t +\"\ \";
}
public String toString(){
return \"From: \"+sender+\"\ \ \"+text+\"\ To: \"+recipient+\"\ \";
}
}.
In which of the following structures are spores formed (Select all .pdf
In which of the following structures are spores formed? (Select all that apply.)ovule conespollen
conesNone of the answer options is correct.carpelsstamensIn which of the following structures
are spores formed? (Select all that apply.)ovule conespollen conesNone of the answer options is
correct.carpelsstamens
Solution
SPORES:it is very light and carried great distances survive in un favourable droughts and
temperatures.A spore is a single cell that develop in to exact copy of plant from which it came.it
is very small,spores core contain cytoplasm and dna.
spores are formed in carpels and stamens.the stamens and carpels of flower contain sprongia,in
which spores are developed..
More Related Content
Similar to A mother and father both have normal karyotypes. The father is color .pdf
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.
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..
Gregor Mendel used pea plants to study heredity in a series of exper.docx
Gregor Mendel used pea plants to study heredity in a series of experiments. Mendel worked by carefully observing and recording traits in successive generations of plants. Knowledge about DNA and chromosomes came later.
This lab will apply genetic laws to human inheritance using Punnett squares.
Recall that DNA is wound tightly into chromosomes. Cells with only one set of chromosomes, such as sex cells, are
haploid
. When two haploid cells fuse during fertilization, a diploid zygote with two full sets of chromosomes is formed. Most cells of a mature individual are diploid.
Homologous chromosomes
have the same genes, but they might have different versions (
alleles
) of those genes.
Diploid
cells have two alleles for each gene. These alleles might be identical (gene A) or different (gene B). Each gene’s
locus
is its location on a chromosome.
Human traits come through dominant or recessive inheritance. For example, the cystic fibrosis traits carried by a dominant allele are always expressed, even if the recessive gene is present (FF or Ff). The recessive is only expressed when two copies of the recessive allele are present (ff).
Mother: Healthy carrier
F
f
Father: Healthy carrier
F
FF
Healthy non-carrier
Ff
Healthy carrier
f
Ff
Healthy carrier
FF
Affected
Human gender is carried on the X and Y chromosomes. Females are XX and males are XY. Heredity traits such as color blindness, which is the inability to distinguish among some colors, are carried on the X chromosome (X
c
). The presence of one normal X
C
will allow normal vision.
In this next set of exercises, you will determine the genotypes of the parents by considering the inheritance patterns of traits in their children. The following is a table of the phenotypes of the family members:
Phenotype
Alleles
Parents
Mother
Not color blind
Freckles
Type B blood
X
c
X
C
Ff
I
B
i
Father
Color blind
No freckles
Type A blood
X
c
Y
Ff
I
A
i
Children
Abby
Color blind
Freckles
X
c
X
c
Ff or ff
Brady
Not color blind
No freckles
X
C
Y
ff
Carly
Not color blind
No freckles
X
c
X
C
ff
Dennis
Color blind
Freckles
X
c
Y
Ff or ff
Exercise 1: Color Blindness
Using the alleles XC (not color blind) and Xc (color blind), distribute the gametes from each parent to the outside of the Punnett square. Drag and drop the child with the correct phenotype to the box within the Punnett square that has the corresponding genotype that would occur from the fusion of egg and sperm as indicated by your placement of the gametes.
Exercise 2: Freckles
Freckles are groups of cells on the skin that produce the pigment melanin, often in response to exposure to ultraviolet (UV) light. The gene for freckles is inherited in a dominant/recessive pattern. A person carrying even a single copy of the dominant allele (F) will have freckles. A person who is homozygous recessive (ff) will have no freckles.
Using the alleles F (freckles) and f (no freckles), distribute the gametes from each parent to the outside of the Punne.
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.
Similar to A mother and father both have normal karyotypes. The father is color .pdf (20)
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
uetator Periodic Table Question 6 of 30 Incorrect Incorrect DA Saplin.pdf
uetator Periodic Table Question 6 of 30 Incorrect Incorrect DA Saplin.pdf
Heredity: inheritance and variation LESSON IN SCIENCE 9
Heredity: inheritance and variation LESSON IN SCIENCE 9
Gregor Mendel used pea plants to study heredity in a series of exper.docx
Gregor Mendel used pea plants to study heredity in a series of exper.docx
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 .pdf
More from connellalykshamesb60
JAVA ProgrammingDesign a class Message that models an e-mail messa.pdf
JAVA Programming
Design a class Message that models an e-mail message. A message has a recipient, a sender, and
a message text. Support the following methods:
a. A constructor that takes the sender and recipient
b. A method append that appends a line of text to the message body
c. A method toString that makes the message into one long string like this: “From: Fname
Lname\ To: Some Recipient\ …”
Write a driver program that uses this class to make a message and print it.
Solution
MessageTest.java
import java.util.Scanner;
public class MessageTest {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter Recipient Name: \");
String recipientName = scan.nextLine();
System.out.println(\"Enter Sender Name: \");
String senderName = scan.nextLine();
System.out.println(\"Enter the body text: \");
String text = scan.nextLine();
Message m = new Message(recipientName, senderName);
m.append(text);
System.out.println(m.toString());
}
}
Message.java
public class Message {
private String recipient;
private String sender;
private String text;
public Message(String r, String s){
this.recipient = r;
this.sender = s;
text = \"\";
}
public void append(String t){
text = text + t +\"\ \";
}
public String toString(){
return \"From: \"+sender+\"\ \ \"+text+\"\ To: \"+recipient+\"\ \";
}
}.
In which of the following structures are spores formed (Select all .pdf
In which of the following structures are spores formed? (Select all that apply.)ovule conespollen
conesNone of the answer options is correct.carpelsstamensIn which of the following structures
are spores formed? (Select all that apply.)ovule conespollen conesNone of the answer options is
correct.carpelsstamens
Solution
SPORES:it is very light and carried great distances survive in un favourable droughts and
temperatures.A spore is a single cell that develop in to exact copy of plant from which it came.it
is very small,spores core contain cytoplasm and dna.
spores are formed in carpels and stamens.the stamens and carpels of flower contain sprongia,in
which spores are developed..
How does Windows managestore user names and passwords How does it .pdf
How does Windows manage/store user names and passwords? How does it differ from Unix?
What privileges does the \"root\" UID have in Linux?
Solution
1) the usernames and passwords in windows are hashed using \"NTLMv2\"algorithm which is a
part of the suite of windows security protocols and finally encrypted credentials are stored in
SAM (security account manager) present in the following directory
C:\\windows\\system32\\config\\SAM
2) the user credential storage of windows differs from unix as said below :
In old unix variants ,passwords are stored in etc/passwd in encrypted form.
But in newer versions ,storage is etc/shadow which ensures no one can access the passwords
except the root.
3)privileges of root in linux :
- ability to modify OS
- grant and revoke permissions
- can access passwords file
- read and modify configuration files.
How to use class set a function(double money(); receive a number and .pdf
How to use class set a function(double money(); receive a number and then output with this
number*2) and use it in the main, cpp?
Solution
//Here number is a part of class definition.
//As your function definition money does not include any number variable i am assuming it to be
part of class.
#include
using namespace std;
class Program
{
private:
double number;
public:
Program()
{
number = 0;
}
Program(int number)
{
this->number = number;
}
double money()
{
return 2*number;
}
};
int main()
{
Program p(10);
cout << p.money() << endl;
return 0;
}.
For each of the following descriptions indicate which best describes.pdf
For each of the following descriptions indicate which best describes proteins or protein
sequences that are:
(S) similar; (H) homolgous; (O) orthologous; (P) paralogs; (D) domains; (F) a superfamily
____ sequences that have a shared evolutionary ancestry
____ sequences that have a certain statistically significant % of identical amino acids when
aligned
____ proteins that have a very high % of identical amino acids when aligned and which are
found in the same
Organism
____ sequences that are small but share structural and functional similarity
____ sequences that have a shared ancestral sequence that changed during a speciation event
A given protein has been monoubiquitylated. What is the expected result?
a. the protein will be activated to turn on transcription
b. the subcellular localization of the protein will change
c. the protein will be activated to repair the DNA
d. the protein will be activated to phosphorylate a substrate
e. any of the above
f. none of the above
The synthetic toxin 2,4-dinitrophenol can uncouple ATP synthesis from mitochondrial
respiration by decreasing the permeability of the inner membrane to protons. What would be the
effect of dinitrophenol treatment on the amount of ATP produced by mitochondria and on the
rate of ATP transport across the inner membrane, respectively?
a. positive, positive
b. positive, negative
c. negative, positive
d. negative, negative
Which of the following requires the mitochondria to create contact points with the ER?
a. mt fission
b. mt fusion
c. mt biogenesis of more mitochondria
d. autophagy of mitochondria (Mitophagy)
e. apoptosis
Solution
S = proteins that have a very high % of identical amino acids when aligned and which are found
in the same Organism
H = sequences that have a shared evolutionary ancestry
O = sequences that have a shared ancestral sequence that changed during a speciation event
P = sequences that are small but share structural and functional similarity
D = Discription not mentioned in the question (A domain is a well-structured 3D, independent
and fully functional part of a protein. Mostly various domains combine to form a proetin to
execute a specific function in cells e.g. enzymes)
F = sequences that have a certain statistically significant % of identical amino acids when
aligned.
Explain how an Annelidian differs from a MolluscanSolutionMoll.pdf
Explain how an Annelidian differs from a Molluscan?
Solution
Mollusks are known to very diverse including 110,000 named species and some examples are
clams, oysters, snails, octopi etc. These mollusks are known to have distant common ancestor
with the annelid worms (trochophore larva) and it has been believed that both mollusks and
annelids most probably might have evolved from free-living flatworms.
The main difference between mollusks and annelids are
Only one heart is seen in Mollusca, while Annelids have five hearts
Open circulatory system is found in Mollusca can have open circulatory system but annelids
have closed circulatory system
Mollusks are marine creatures while Nematoda and Annelids are known to inhabit various
ecologies, marine (tidal zone and hydrothermal vents), fresh water and terrestrial.
Mollusks have a more complex system when compared to annelids
Either externally or internally we can spot shells in Mollusks (snails, scallops, and squids), but
shells are absent in Annelids..
DNA s composed of nucleotides that are joined together by covalent bo.pdf
DNA s composed of nucleotides that are joined together by covalent bonds between which two
parts of each nucleotide? Between two sugars Between the sugar and the organic base.
Between the sugar and the phosphate group. Between the phosphate group and the organic base.
Solution
DNA is composed of nucleotides that are joined together by covalent bonds between which wo
parts of each nucleotide:
Betweeen the sugar and phosphate group.......................
Discuss some of the challenges involved in ensuring that critical kn.pdf
Discuss some of the challenges involved in ensuring that critical knowledge is transferred both to
present-day employees as well as future ones. (1 ~ 2 Paragraphs)
Solution
The challenges involved in ensuring both present and future employees are of many kinds.
Ensuring critical knowledge, the main is to provide security . Providing the right level of security
is the main part and shouldnot access from other systems. Employees come from different places
, cultures, and habits making them united in the office for working together can be a challenge.
The next comes, keeping up with technology , in which the knowledge could be shared to others
as easy and faster.Constant changing of the structures mean learning how to be smart and quick.
Interpretation effectively is that acquiring the correct and valuable information with in time
which can be used by other departments. Here comes the, last main one i.e., impressing the
person who works actively and effectively . By doing these kind of works not only the purticular
employee but also other persons can be given a way for encouraing the work .
To overcome the knowledge challenge facilities , can be shared by data bases and technological
aids.
Knowledge management contributes towards streamlining the ideas problems, projects and
deployment driving towards productivity.
But, it’s more than just knowing everything . It’s creating a synthesis between the people and the
information to the point that the whole is more than the sum of the parts..
Biologically, the term is synonymous with [Choose the term race. refe.pdf
Biologically, the term is synonymous with [Choose the term race. refers to the position of a
particular gene Choose on a chromosome. Traits that provide a benefit in particular [Choose
environments typically create when their prevalence within a population is mapped onto the
landscape. If two different alleles phenotypically result in a trait [Choose J that appears to be
blended, the alleles are said to exhibit can lead to speciation events. Choose s one of the four
fields of anthropology. Choose The structure of DNA was first discovered by [Choose Francis
Crick, Rosalind Franklin, and A type blood and B type blood are Both Choose appear in the
phenotype if they are in the genotype. states that only one of the two alleles (choose l present in
the parent will be present in the gamete produced by that parent.
Solution
1. Clines
2. Locus
3. Subspecies
4. Incomplete dominance
5. Disruptive selection
6. Linguistics
7. James Watson
8. Codominant
9. Law of segregation.
Anyone good with linguistics Are the following headlines transitive.pdf
Anyone good with linguistics? Are the following headlines transitive or intransitive?
\"Clinton Says Inaction By Russia May Lead to Syrian Civil War\"
\"Russian Church Is a Strong Voice In Opposing Intervention in Syria\"
\"Putin Reject Intervention But Fears Civil War In Syria\"
\"Syria: Heavy Clashes Reported\"
\"Syria Bars 17 Western Diplomats and Allows Increased Aid Agency Presence\"
Solution
Let us look at the dictionary meaning of transitive and intransitive.
Transitive verb 1. to lead astray or deceive; especially : SEDUCE 2 : to deliver to an enemy by
treachery 3 : to fail or desert especially in time of need 4 a : to reveal unintentionally b : SHOW,
INDICATE c : to disclose in violation of confidence intransitive verb : to prove false.
Intransitive verb is to practice deceit; also : to give a false impression
Looking at the headlines
\"Clinton Says Inaction By Russia May Lead to Syrian Civil War\" - TRANSITIVE
\"Russian Church Is a Strong Voice In Opposing Intervention in Syria\" - INTRANSITIVE
\"Putin Reject Intervention But Fears Civil War In Syria\" - TRANSISITVE
\"Syria: Heavy Clashes Reported\" - INTRANSITIVE
\"Syria Bars 17 Western Diplomats and Allows Increased Aid Agency Presence\" -
TRANSITIVE.
A man has a condition where all of his gametes undergo nondisjunctio.pdf
A man has a condition where all of his gametes undergo nondisjunction of the sex chromosomes
in meiosis I, but meiosis II proceeds normally. He mates with a woman who produces all normal
gametes. What is the probability that the fertilized egg will develop into a child with Klinefelter
syndrome (XXY)? Assume that all gametes and zygotes are viable.
1/8
1/4
1/3
1/2
0
1/8
1/4
1/3
1/2
0
Solution
Answer: 1/2
Nondisjunction occurs when homologous chromosomes, in this case the X and Y or two X sex
chromosomes, fail to separate, producing a sperm with an X and a Y chromosome. If the man
has nondisjunction of all of his sperms (at meiosis1), then 50% or ½ of the sperms are XY.
Therefore probability of child with Klinefelter syndrome (XXY) will be 1/2
==.
____ The pigment produced by S. marvel scenes would appear first on t.pdf
____ The pigment produced by S. marvel scenes would appear first on the edge of colony
because these bacteria reach stationary phase more quickly than those in the middle. ____
Typhoid Mary was considered to have an acute infection with the organism that causes
tuberculosis; her gall bladder, where bile is made, contained the organism.
Solution
1. True, this organism produce red pigmentation peviously this organism was considered as non
pathogenic organism but now some harmfull activity are also known
2.True, the tuberculosis in the abdominal part is most common type of extra pulmonary
tuberculosis.
Write the advantages of using JavaFX with compare to Swing and AWT..pdf
Write the advantages of using JavaFX with compare to Swing and AWT.
Solution
The advantages of using java fx with compared to swing and awt are:
1.The first advantage is that java fx works similarly to a pdf.The java fx is basically a tree/node
struction which has various values that are inherited from the nodes and the swing is basically
iterative and has a flat structure .The awt is similar to swing as well.The PDF is built around a
tree structures which makes it easier to map onto JavaFx as compared to swing.It can also handle
xforms.
2.The second advantage is that JavaFx has more features as compared to swing and awt.For
example in order to create the shapes it is easier using javafx with variable line as compared to
swings.
3.The third advantage is JavaFx uses a more compact and a simpler code.The code is short as
compared to swing.JavaFx takes low level tedium out of the code which means we do not have
to implement paint() and print() methods in javafx which we have to do in swing.
4.It is an integral part of java8 and a lot of focus is going on to make it compatible and run on
android devices and rasberry pi operating systems.It can also interact with the existign swing
applications.It can be run on various devices.
5.The new developments in java will take place in javafx and it will make it user friendly with
easier coding..
Write a program in c++ that produces a bar chart showing the populat.pdf
Write a program in c++ that produces a bar chart showing the population growth of Prairieville, a
small town in the Midwest, at 20-year intervals during the past 100 years.
The program will take input from user for population figures (rounded to the nearest 1,000
people) for 1900, 1920, 1940, 1960, 1980, and 2000. For each year it should display the date and
a bar consisting of one asterisk for each 1,000 people.
Here is an example of how the chart might begin:
PRAIRIEVILLE POPULATION GROWTH
(each * represents 1,000 people)
1900 **
1920 ****
1940 *****
After the bar chart, print out the maximum, minimum and average number of population.
Solution
#include
using namespace std;
int main()
{
int year[6]={1900,1920,1940,1960,1980,2000};
long population[6],calc[6];
int i,j;
cout<<\"Enter the population for years\";
for(i=0;i<6;i++)
{
cin>>population[i]; //input years
}
cout << endl << \"PRAIRIEVILLE POPULATION GROWTH\";
cout << endl << \"(each * represents 1,000 people) \" << endl;
for(i=0;i<6;i++)
{
calc[i]= population[i]/1000; //calcualte population per 1000 persons
}
for(i=0;i<6;i++)
{
cout << \"\ \"<.
Write a 2 page essay describing what life was like in the Great Depr.pdf
Write a 2 page essay describing what life was like in the Great Depression 1929-1941 with a
focus on the important events and data which characterized economic conditions of that time
period. Cite at lease 4 outside sources of supporting information/data from the internet.
Solution
CAUSES
EFFECTS.
Why do silver nanoparticles of 10 nanometers in diameter have lower m.pdf
Why do silver nanoparticles of 10 nanometers in diameter have lower melting temperature than
the same silver but in the form of a 1/4 inch rod?
Solution
Firstly in case of spherical there is low volume to surface area so it has high heat absorbing
capacity so low melting point but in case of cylindrical surface to volume ratio is quite large
Secondly in case of spherical nanoparticle browniam motion is very high because of which high
heat transfer coeeficient is achieved and so low melting point in case of nanoparticle.
Why Computer System Management is so critical for the companies (50.pdf
Why Computer System Management is so critical for the companies? (500-600 words)
Solution
Managing computer systems is very difficult for the company so it appoints managers for it who
have the following responsibilities to carry on:
Computer and information systems managers, often called information technology (IT)
managers or IT project managers, plan, coordinate, and direct computer-related activities in an
organization. They help determine the information technology goals of an organization and are
responsible for implementing computer systems to meet those goals.
Duties:
Computer and information systems managers typically do the following:
Analyze their organization’s computer needs and recommend possible upgrades for top
executives to consider.
Plan and direct the installation and maintenance of computer hardware and software.
Ensure the security of an organization’s network and electronic documents.
Assess the costs and benefits of new projects and justify funding on projects to top executives.
Learn about new technology and look for ways to upgrade their organization’s computer
systems.
Determine short- and long-term personnel needs for their department.
Plan and direct the work of other IT professionals, including computer systems analysts,
software developers,information security analysts, andcomputer support specialists.
Negotiate with vendors to get the highest level of service for their organization’s technology
Few managers carry out all of these duties. There are various types of computer and information
systems managers, and the specific duties of each are determined by the size and structure of the
firm. Smaller firms may not employ every type of manager.
The following are examples of types of computer and information systems managers:
Chief information officers (CIOs) are responsible for the overall technology strategy of their
organizations.
They help determine the technology or information goals of an organization and then oversee
implementation of technology to meet those goals.
CIOs may focus on a specific area, such as electronic data processing or information systems,
but CIOs tend to focus more on long-term or big picture issues. At small organizations a CIO has
more direct control over the IT department, and at larger organizations other managers under the
CIO may handle the day-to-day activities of the IT department.
CIOs who do not have technical expertise and who focus solely on a company’s business aspects
are included in a separate profile on top executives.
Chief technology officers (CTOs) evaluate new technology and determine how it can help their
organization.
When both CIOs and CTOs are present, the CTO usually has more technical expertise.
The CTO is responsible for designing and recommending the appropriate technology solutions to
support the policies and directives issued by the CIO. CTOs also work with different
departments to implement the organization’s technology plans.
The CTO usually reports directl.
Which of the following statements is not true about facilitated diff.pdf
Which of the following statements is not true about facilitated diffusion?
a. It requires specific transport molecules in the plasma membrane.
b. A transporter protein undergoes a change in shape to accomplish the transfer.
c.A substrate molecule is transported from an area of high concentration to low concentration.
d. It requires ATP.
e. It is driven by osmotic pressure.
Solution
B. A transporter protein undergoes a change in shape to accomplish the transfer.
Facilitated diffusion is the process of spontaneous passive transport of molecules or ions across a
cell\'s membrane via specific transmembrane integral proteins. Larger molecules are transported
by transmembrane carrier proteins, such as permeases, that change their conformation as the
molecules are carried across (e.g. glucose or amino acids)..
Which is not a hypothesis for the origin of virusesThey originated .pdf
Which is not a hypothesis for the origin of viruses?They originated in plasmids and transposable
elements.They originated in symbiotic bacteria.They originated with the origin of life.They
originated from cell membrane proteins.
Solution
Anwer: They originated from cell membrane proteins
Reason:
Viruses might have not originated from cell membrane proteins because according virus-first
hypothesis enzyme that synthesizes cell wall or cell membrane may generated a virus first. The
other hypothesis are that viruses might have originated in plasmids and transposable elements
(progressive hypothesis of viral evolution) or originated in symbiotic bacteria or originated with
the origin of life..
1. When does a corporation need a board of directors Who should and.pdf
1. When does a corporation need a board of directors? Who should and should not serve on the
board of directors?
2. Is there a link between corporate governance and social responsibility? Explain.
3. How can a decision maker identify strategic factors in the corporation\'s external
environment? How will you do this in your case analysis?
4. What is the relevance of the resource-based view of a firm to strategic management in a global
environment?
Solution
1. Corporation needa board of directors when it does not have all the skills and expertise to run
its operations effectively. The members of the board bring in a lot to the table in terms of their
experience and skills in matters related to law, strategy, finance and marketing etc. This helps the
company grow in an effective manner.
Board of directors consists of internal members of an organization and independent directors.
Internal members should usually consist of CEO and top management members. Employees in
the low rung of hierarchy should not serve on the board. Independent directors should not be a
person who has a financial interest in the company.
2. Yes, there is a link between corporate governance and social responsibility. Corporate
governance looks at maintaining the balance between the economic and social goals of a
company. It helps in streamlining the interest of the company with that of the society.
CSR or corporate social responsibility aims at ethical treatment of stakeholders. It aims at
conducting the company in a socially responsible manner.
Having corporate governence in a company encourages the implementation of CSR activities. A
company with poor corporate governance will not be doing much in terms of CSR initiatives.
So there is a direct relation between corporate governance and CSR..
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