Why is mild to moderate expression of an X-linked recessive trait no.pdf
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Why is mild to moderate expression of an X-linked recessive trait not observed in heterozygous female Drosophila? Solution In X-linked recessive trait, females will get the mutant phenotype if they receive both copies of the defective gene or females who are homozygous for the gene mutation. If they have one dominant and one recessive allele, they will not show any mutant phenotype. SSex-linkedgenes are located on the on the X chromosomes, for example wild type of drosophila will have red eyes, and white color eyes are recessive. If a female drosophila possesses XRXr(heterozygous condition), drosophila eyes will be red in color, if gets both the recessive alleles XrXr(homozygous condition), then only her eyes will become white. That is the reason why heterozygous female drosophila will not express the mild to moderate X-linked recessive trait. Due to the presence of two copies of X chromosomes, only one copy of wild allele is enough for them to get normal phenotype..
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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.
sex linked inheritance, Sex Influence inheritance and sex limited characters
sex linked inheritance, Sex Influence inheritance and sex limited characters
Recommended
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.
sex linked inheritance, Sex Influence inheritance and sex limited characters
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This presentation intends to explore the sex-linked characters along with some fatal diseases of human beings, their cause, consequences and other issues.
Heritable Traits in Man, Pedigree Analysis and Pedigree Application
This presentation is required in our Genetics Subject
Sex linked inheritance
Concept of Sex chromosomes and autosomes,
Inheritance of X- linked genes – eye colour in Drosophila,
Inheritance of colour blindness in humans,
Inheritance of Y-linked Genes -Holandric genes in humans,
Sex influenced genes – baldness in humans
Sex-limited genes - feathering in domestic fowl
Allelic and non allelic interactions & Drosophila mutants.pptx
Allelic and non allelic interactions & Drosophila mutants
not exaclty sure what to do here U Sexlinked Disorders Se.pdf
not exaclty sure what to do here
U. Sex-linked Disorders Sex-linked traits ine controlled by genes that are found on the X
clromosome, but not on the Y chromosome in humans. The X chromosome is larger and has more
genes than the Y chromosome. This complicntes the inleritance of uraits on the X clromosome
since we expect that a normal diploid cell will have two alleles for any gene. Such is not he case
for males siuce they are XY and will only have cne copy of the genes that are locied ou the X
chromosome: Females (XX) will have two ulleles for a gene located on the X chromosome Males (
XY) will only have one allele for that trait. This, females with one dominant and one recessive
allele for a gene on the X chromosene would display the dominant phenotype. However a mate
will always displey the phenorype of the one allele that they have for the mait that is carried on the
X chroinosonie. A new way of writing alleles: XR - normal eyesiglt X-colorblind Genotypes: XRXR
XRXrXRYXRXrXY Phenotypes: Normal Eyesight and Colorblind Remember to follow the same
steps in order to solve problems: 1. Assign lefters to represent the genes 2. Determine the
parental phenotypes and genotypes. 3. Predict the outcome of a cross between the possible
gameles from the tiro parents. 4. State plenotypic and genoty pic ratios present ia the offspring. 5.
Don't forget to answer the question witlin the problem. ains. begin{tabular}{|c|c|} hlinex & x hlinexn
x & xx hlinexy & xy hline end{tabular}.
INHERTANCE PATTEREN IN ANMALS,SEGREGATION ,INDEPENDENT ASSORTMENT,MULTIPLE AL...
MILLER AND HARLEY 5TH EDITION
CHAPTER # 3 "CELL DIVISION AND INHERITANCE"
TOPIC : INHERITANCE PATTERN IN ANIMALS "
Sex-linked inheritance by Puzon and Tope
Sex-linked inheritance by Puzon, kathleen Faye R. and Tope, Donita J.
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.
In general diffusion is faster forA. less closely packed structure.pdf
In general diffusion is faster for
A. less closely packed structure
B. higher melting points solids
C. substitutional relative to interstitial diffusion
D. all of the above
Solution
DIFFUSION FASTER FOR
- open crystal structures
- lower melting point temperature materials
- smaller diffusion atoms
- lower density materials
- cations
DIFFUSION IS SLOWER FOR
- close packed structures
-higher melting point temp materials
- anions
- high density materials
therefore option A is correct
substitutional diffusion is much slower than that of interstitial.
In your own words 1.Explain what a semiconductor is. 2.Describe one.pdf
In your own words: 1.Explain what a semiconductor is. 2.Describe one aspect of the operation of
a semiconductor using principles from chemistry (for example: intrinsic, extrinsic, unbiased,
forward bias, or reverse bias). 3.Give one application of semiconductors in technology.
Solution
The materials in which the valance band is fully filled, conduction band is almost empty and
forbidden band is very small about 1ev,are considered as semiconductors
Silico and germanium crystal in its purest form is considered as intrinsic semiconductors.
N_type si+p(Silicon +phosphorus)
P_type b+si (boron+silicon)
Diodes, mosfets.
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Similar to Why is mild to moderate expression of an X-linked recessive trait no.pdf
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.
Heritable Traits in Man, Pedigree Analysis and Pedigree Application
This presentation is required in our Genetics Subject
Sex linked inheritance
Concept of Sex chromosomes and autosomes,
Inheritance of X- linked genes – eye colour in Drosophila,
Inheritance of colour blindness in humans,
Inheritance of Y-linked Genes -Holandric genes in humans,
Sex influenced genes – baldness in humans
Sex-limited genes - feathering in domestic fowl
Allelic and non allelic interactions & Drosophila mutants.pptx
Allelic and non allelic interactions & Drosophila mutants
not exaclty sure what to do here U Sexlinked Disorders Se.pdf
not exaclty sure what to do here
U. Sex-linked Disorders Sex-linked traits ine controlled by genes that are found on the X
clromosome, but not on the Y chromosome in humans. The X chromosome is larger and has more
genes than the Y chromosome. This complicntes the inleritance of uraits on the X clromosome
since we expect that a normal diploid cell will have two alleles for any gene. Such is not he case
for males siuce they are XY and will only have cne copy of the genes that are locied ou the X
chromosome: Females (XX) will have two ulleles for a gene located on the X chromosome Males (
XY) will only have one allele for that trait. This, females with one dominant and one recessive
allele for a gene on the X chromosene would display the dominant phenotype. However a mate
will always displey the phenorype of the one allele that they have for the mait that is carried on the
X chroinosonie. A new way of writing alleles: XR - normal eyesiglt X-colorblind Genotypes: XRXR
XRXrXRYXRXrXY Phenotypes: Normal Eyesight and Colorblind Remember to follow the same
steps in order to solve problems: 1. Assign lefters to represent the genes 2. Determine the
parental phenotypes and genotypes. 3. Predict the outcome of a cross between the possible
gameles from the tiro parents. 4. State plenotypic and genoty pic ratios present ia the offspring. 5.
Don't forget to answer the question witlin the problem. ains. begin{tabular}{|c|c|} hlinex & x hlinexn
x & xx hlinexy & xy hline end{tabular}.
INHERTANCE PATTEREN IN ANMALS,SEGREGATION ,INDEPENDENT ASSORTMENT,MULTIPLE AL...
MILLER AND HARLEY 5TH EDITION
CHAPTER # 3 "CELL DIVISION AND INHERITANCE"
TOPIC : INHERITANCE PATTERN IN ANIMALS "
Sex-linked inheritance by Puzon and Tope
Sex-linked inheritance by Puzon, kathleen Faye R. and Tope, Donita J.
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.
Similar to Why is mild to moderate expression of an X-linked recessive trait no.pdf (20)
Heritable Traits in Man, Pedigree Analysis and Pedigree Application
Heritable Traits in Man, Pedigree Analysis and Pedigree Application
Allelic and non allelic interactions & Drosophila mutants.pptx
Allelic and non allelic interactions & Drosophila mutants.pptx
not exaclty sure what to do here U Sexlinked Disorders Se.pdf
not exaclty sure what to do here U Sexlinked Disorders Se.pdf
INHERTANCE PATTEREN IN ANMALS,SEGREGATION ,INDEPENDENT ASSORTMENT,MULTIPLE AL...
INHERTANCE PATTEREN IN ANMALS,SEGREGATION ,INDEPENDENT ASSORTMENT,MULTIPLE AL...
Consider the following pedigree, which traces the inheritance of a si.pdf
Consider the following pedigree, which traces the inheritance of a si.pdf
More from pasqualealvarez467
In general diffusion is faster forA. less closely packed structure.pdf
In general diffusion is faster for
A. less closely packed structure
B. higher melting points solids
C. substitutional relative to interstitial diffusion
D. all of the above
Solution
DIFFUSION FASTER FOR
- open crystal structures
- lower melting point temperature materials
- smaller diffusion atoms
- lower density materials
- cations
DIFFUSION IS SLOWER FOR
- close packed structures
-higher melting point temp materials
- anions
- high density materials
therefore option A is correct
substitutional diffusion is much slower than that of interstitial.
In your own words 1.Explain what a semiconductor is. 2.Describe one.pdf
In your own words: 1.Explain what a semiconductor is. 2.Describe one aspect of the operation of
a semiconductor using principles from chemistry (for example: intrinsic, extrinsic, unbiased,
forward bias, or reverse bias). 3.Give one application of semiconductors in technology.
Solution
The materials in which the valance band is fully filled, conduction band is almost empty and
forbidden band is very small about 1ev,are considered as semiconductors
Silico and germanium crystal in its purest form is considered as intrinsic semiconductors.
N_type si+p(Silicon +phosphorus)
P_type b+si (boron+silicon)
Diodes, mosfets.
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If the random variable X is normally distributed with mean and standard deviation , then the
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True.
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I have a Computer Archtecture book and notes that describe addressin.pdf
I have a Computer Archtecture book and notes that describe addressing modes. Specifically
those pertaning to MIPS Assembler. I know there are 5 modes MIPS supports but when I look at
the book I see some named differently to what I see in the notes. As someone learning I need
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Solution
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register no alternate name
base displacement,base-offset
the next 2 come in similaar category
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the next 3 come in similar category
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I really need help with this Assignment Please in C programming not .pdf
I really need help with this Assignment Please in C programming not C++ or C#, just C. Thank
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Hello , Choose please one of the following program types:
The assignment is to write a menu driven program that manages a small business inventory,
collection or list. You pick the topic. The focus of your program is up to you.
The Menu commands must include:
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A....Add a new entry
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Q...Quit
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the inventory to be modified. i.e. add inventory quantities, change price/cost, change dates, etc.
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Suggestions include: item name, manufacturer, etc o Minimum of 2 integers – 1 must be product
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When you add new item the program will ask the user for each of the fields on a separate line.
When you delete an item from inventory the program will ask you for the integer id of the entry
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location of the deleted entry
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labeled and displayed.
Creating a current inventory report copies the current entries in the array to an output file. This
must include labeling all of the information so that it is clear what information is being provided.
Clearing the records deletes all of the information in the array.
When you add new item the program will ask the user for each of the fields on a separate line.
When you delete a question from the list, the program will ask you for the question id of the
entry to be deleted, locate the entry in the array and remove all of the data for that entry. – The
list does not need to be sorted – to remove an entry, you may move the last item in the list to the
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When you play the game, you must display the question, the possible answers and the point
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incorrectly, the correct answer will be displayed. Creating a current report copies the current
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so that it is clear what information is being provided.
Clearing all questions deletes all of the information in the array.
Instructions:
You should use at least 10 user-defined functions (plus main) to appropriately break the problem
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Identify opportunities for leadership for sustainability in personal.pdf
Identify opportunities for leadership for sustainability in personal, professional, and community
life?
Solution
1> socialising: It tends to increase public relationships and support from the society which helps
in building confidence in an individual.
2> family relationships: It tends to create a strong emotional bond which leads to care and
attention. It helps in developing leadership quality.
3> team work: In any profession team work is required to accomplish a succesful task. It helps in
building leadership quality..
Gametes that contain new combinations of alleles not present in the .pdf
Gametes that contain new combinations of alleles not present in the parents are called _____
gametes
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B. haplotype
C. repulsion
D. recombinant
E. restricted
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Gamete is a mature haploid male or female germ cell which is able to unite with another of
opposite sex in the process of sexual reproduction. Recombinant gametes are the gametes which
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From Figure 26 what is true of Dryopithecus and Ouranopithecus 1.) .pdf
From Figure 26 what is true of Dryopithecus and Ouranopithecus? 1.) They were great apes.
2.) They shared a common ancestor more recently with the orangutans than with the other great
apes. 3.) They appear to be part of a polytomy. 4.) Their closest common ancestor with all of
the extant great apes is the one they share with the orangutans. 5.) They were about as different
from each other genetically as humans are different from chimps. A. 1, 2, and 4 B. 1, 3, and 4
C. 1, 2, 3, and 4 D. 2, 3, 4, and 5 Which of these can be properly inferred from the phylogeny in
Figure 26-5? A. Chimps and humans evolved from gorillas. B. The lesser apes are genetically
more distinct from each other than the members of the great apes are each other. C. Orangutans
have existed for about 14 million years. D. Together, the lesser apes and great apes form a clade.
Assuming chimps and gorillas are closest relatives to humans, removing humans from the great
ape clade and placing them in a different clade has the effect of making the phylogenetic tree of
the great apes A) polyphyletic. B) paraphyletic. C) monophyletic. D) aphyletic
Solution
they share 4.1 .they are not great apes.2 is definitely true, they share a common ancestor more
recently with the orangutans than the other apes.3. We can see 2 branch of dryopithecus &
ouranopothecus rising from same point. So they are a part of polytomy 4 is also true their closest
common ancestor with all of the extant great apes is the one they share with orangutan. 5. Is true
they are obviously genetically different as we see they are part of different branches. So , last
option is right I.e. 2,3,4,5 are correct
5. It is clearly shown that orangutans existed for 14 million years.
6..
F. Chlorine Content F1. What type of sample might be expected to cont.pdf
F. Chlorine Content F1. What type of sample might be expected to contain chlorine? F2. What
are the units for chlorine? F3. What range of concentrations of chlorine is be typical in treated
drinking water (look that up?)? Is our test sensitive enough to measure the chlorine that might be
found in drinking water? F4. Take three Chlorine measurements for each sample, and average
the values. Why? F5. What does the Chlorine tell you about the potential origin of each sample?
Solution
1.Chlorine is one among the top 20 elements abudantly available on the earth. It is present both
in earth crust and mostly in sea water. The type of samples mostly contains chlorine are in
seawater it is available as mostly in NACl, KCl. On earth crust it is available as halite, or rock
salt (NaCl), sylvite (KCl), and camallite (KCl MgCl 2 ).
2. Concentration of any of the elements are mostly measured in the units ppm (parts per million).
3.The accpetable limit of chlorine (as Cl) is 250 mg/l or 250 ppm. In absence of any other
soluble source 1000 ppm where as the limit ( Cl2) is 4 ppm. The tests are sensitive enough to
measure chlorine content . Using colorimeter devices the quality of the water is measured with
that of the stds. this is the most ease way.
4. The avg of the measurments is taken to minimize the difference between initial and final
values. The conditions keep changng every instant as the chlorine get diluting in water and also
because of the atm conditions the concentration of chlorine may change. So we go for an avg
value.
5.The potential origin refers to the content of chlorine..
Describe each level of cache. Time sharing systems do what for the u.pdf
Describe each level of cache. Time sharing systems do what for the user? and What is the
fundamental difference between multitasking and multiprocessing environments?
Solution
1)
Cache memory is fast and expensive. Traditionally, it is categorized as \"levels\" that describe its
closeness and accessibility to the microprocessor:
Level 1 (L1) cache is extremely fast but relatively small, and is usually embedded in the
processor chip (CPU).
Level 2 (L2) cache is often more capacious than L1; it may be located on the CPU or on a
separate chip or coprocessor with a high-speed alternative system bus interconnecting the cache
to the CPU, so as not to be slowed by traffic on the main system bus.
Level 3 (L3) cache is typically specialized memory that works to improve the performance of
L1 and L2. It can be significantly slower than L1 or L2, but is usually double the speed of RAM.
In the case of multicore processors, each core may have its own dedicated L1 and L2 cache, but
share a common L3 cache. When an instruction is referenced in the L3 cache, it is typically
elevated to a higher tier cache.
2)
Time sharing is a technique which enables many people, located at various terminals, to use a
particular computer system at the same time. Time-sharing or multitasking is a logical extension
of multi-programming. Processor\'s time which is shared among multiple users simultaneously is
termed as time-sharing.Time-Sharing Systems objective is to minimize response time. Time
sharing refers to the allocation of computer resources in time slots to several programs
simultaneously.
eg: mainframe computer having many users logged in
Multi-user is a term that defines an operating system that allows concurrent access by multiple
users of a computer.
eg: batch processing systems for mainframe
Time-sharing systems are multi-user systems.
Advantages of Timesharing operating systems are following
Provide advantage of quick response.
Avoids duplication of software.
Reduces CPU idle time.
Multiprocessing - A computer using more than one CPU at a time
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multiprogramming.It means running several tasks on computer.Switching between the jobs is
so fast that the user can interact with the system..
Clearly explain why 2D surface measurement can be unrealisticSo.pdf
Clearly explain why 2D surface measurement can be unrealistic?
Solution
2D and 3D refer to the actual dimensions in a computer workspace. 2D is \"flat\", using the
horizontal and vertical (X and Y) dimensions, the image has only two dimensions and if turned
to the side becomes a line. 3D adds the depth (Z) dimension. This third dimension allows for
rotation and visualization from multiple perspectives. It is essentially the difference between a
photo and a sculpture.
The basic task for measurement is to obtain information about its shape and texture , but all the
things in this real world are 3D but the 2D measurement fails to give information about the
thickness of the surface, and hence cannot be used in the voumetric measurement of objects and
things..
c++ early objectsc++ early objectsc++ early objectsSol.pdf
c++ early objects
c++ early objects
c++ early objects
Solution
Early objects are nothing but the basic concepts the fundational concepts of the programming
language.
There are called \'late\' and \'early\' objects. When someone is learnning a programming language
on their own they start with early objects. Late objects are the deeper concepts like data
structures.
C++ early objects are:
ECT,. are the basic, foundational and must known concepts..
As part of an IT survey, both staff and managers were asked to indic.pdf
As part of an IT survey, both staff and managers were asked to indicate what business and
technical skills are critical to their jobs. In total, 12 skills were included in the survey. One of the
12 skills was “Analyzing Data”. The following table reports findings about that skill:
A. What is the probability that a person selected at random is a manager?
B. What is the probability that a randomly selected person says Analyzing Data is critical skill?
C. What is the probability a randomly selected person is a manager and believes Analyzing Data
is a critical skill?
D. What is the probability that a randomly selected person is a manager or believes Analyzing
Data is a critical skill?
E. Given that a person is a manager, what is the probability that he/she believes Analyzing Data
is a critical skill?
F. The respondents were asked to consider the 12 skills included in the study and pick 4 which
they thought are critical to their job. How many different ways are possible for an individual to
make his/her selection?
G. The respondents were asked to consider the 12 skills included in the study, and rank (as 1, 2,
3, and 4) their top 4 choices. How many different ways are there that a person could make his/her
selection?
Solution
A) PROBABILITY THAT THE PERSON BE A MANAGER = 6264/14074( TOTAL
MANAGER / TOTAL STAFF)= 0.445
B)PROBABILITY THAT THE PERSON SAY IT A CRITICAL SKILL = TOTAL STAFF
WHO SAID SO / TOTAL STAFF = 8007/14074 = 0.568
C)PERSON SELECTED IS MANAGER ANS SAY IT A CRITICAL SKILL = TOTAL
PERSON WHO ARE MANAGER WHO SAY SO / TOTAL STAFF = 3633/14074= 0.258
D)probability that a randomly selected person is a manager or believes Analyzing Datais a
critical skill = HERE OR IS GIVEN THEREFORE WE WILL ADD PEOPLE WHO ARE
MANAGER PLUS PEOPLE WHO BELIEVE THAT IT IS A CRITICAL SKILL/ TOTAL
STAFF = 10638/14074=0.755
Note: we are suppose to answer till 4 subparts for rest repsot again..
a. High Frequency components delayedb. Low frequency components lo.pdf
a. High Frequency components delayed
b. Low frequency components lost or suppressed
c. High Frequency components lost or suppressed
d. Low Frequency components delayed A half-rectified sine wave has been distorted by a poor
measurement system. What type of distortion has resulted in the waveform shown below? 1251
0.7 0.2 0.25 0.75 1.25
Solution
high frequency components delayed.
A signal y(t) is periodic with fundamental period T0, and is the sum.pdf
A signal y(t) is periodic with fundamental period T0, and is the sum of two other signals y(t) =
x1(t) + x2(t). Must x1(t) and x2(t) both be periodic?
Solution
The sum of two periodic signals x1,x2 with periods T1,T2 is periodic only if the ratio of T1/T2 is
a rational number or ratio of 2 integers.X1,X2 must be periodic.To is the LCM of T1,T2..
A die is thrown. Let E be the event that the number thrown is even a.pdf
A die is thrown. Let E be the event that the number thrown is even and F the event that the
number thrown is greater than 4.
Find
(a) E
(b) F
(c) E ? F
(d) E \'
(e) F \'
( f ) E ? F
(g) E - F
(h) E \' ? F
( i ) (E ? F \' ) \' A die is thrown. Let E be the event that the number thrown is even and F the
event that the number thrown is greater than 4. Find (a) Ri (d) E (g) E-F (b) F (e) F (h) EUF (c)
EnF (f) EuF (i) (EnF)
Solution
A die is thrown. Let E be the event that the number thrown is even and F the event that the
number thrown is greater than 4.
Find
(a) E = 2,4,6
(b) F = 5,6
(c) E ? F = 6
(d) E \' = 1,3,5
(e) F \' = 1,2,3,4,
( f ) E ? F = 2,4,5,6
(g) E - F = 2,4,
(h) E \' ? F = 5
( i ) (E ? F \' ) \' = 1,3,5,6.
4. A gaseous fuel contains the following by mole 31.3 C2H6; 67.6 .pdf
4. A gaseous fuel contains the following by mole: 31.3% C2H6; 67.6% CH4; 1.1% N2.
Determine
i. the gravitational stoichiometric air.
ii. The DRY molar composition of the combustion products if the fuel is burned with 150%
theoretical air. Assume complete combustion of the fuel
iii. The WET analysis of the products by mass if the fuel is burned with 15% less air.
Solution
2C2H6 + 7O2 = 2CO2 + 3H2O------ eqn i
0.313 moles of C2H6 + 1.0955 moles of O2 = 0626 moles of CO2 +0.939 moles of H20(here %
of moles is taken as fractions )
given 31.3% of ethane in fuel so moles of O2 = 7/2 *(0.313)= 1.0955
moles of CO2 = 2*0.313 = 0.626
moles of H2O = 3*0.313 = 0.939
CH4+ 2 O2 = CO2 + H2O ------- eqn ii
0.676 moles of methane + 1.352 moles of oxygen = 0.676 moles of CO2 + 0.676 moles of H2O
moles of O2 = 0.676*2 = 1.352
assume air composition
oxygen = 21% = 0.21
nitrogen = 79% = 0.79
moleculae weight of air = 29 g
Total moles of oxygen required for reaction = 1.0955 + 1.352 = 2.4474
= 2.4475 * 16 = 39.16 g ( 1 mole O2 = 16g )
quantity of air needed = 39.16/0.21 = 186.476 g ( in air 21% of O2 is present )
no of moles of air required to burn the enitre fuel = 186.476 / 29 = 6.4137 . ( Gravitational
stoichiometric air )
at dry condition 150% of air is supplied so net moles of air = 6.4137*1.5 = 9.6206
therefore
total moles of CO2 = 0.626+0.676 =1.302 from eqn i & ii
no of moles of nitrogen in 9.6206 moles of air = (9.6206*29 *0.79)/14 =15.74 (molecular weight
of nitrogen = 14 g)
total moles of N2 = 0.011+15.7434 = 15.7534
total quanitity of O2 = (9.6206-6.4137)*29*0.21 =19.53
no of moles of oxygen = 19.53/16 = 1.2206
so Dry molar composition of the combustion products is
%moles of CO2 = 1.302 / ( 1.302+15.7534+1.2206) = 1.302 / 18.276 = 0.0712
%moles of N2 = 15.7534 / ( 1.302+15.7534+1.2206) = 15.7534/18.276 = 0.8619
%moles of O2 = 1.2206/ ( 1.302+15.7534+1.2206) = 1.2206/18.276 =0.06678
At Wet analysis 15% less air is supplied
so total moles of air = 6.4137*0.85 = 5.4516
Wet molar composition of products
moles ofN2 = 0.011
moles of of CO2 = (0.626+0.676)*0.85 = 1.01067 from eqn i &ii
moles of H2O = (0.939+0.676)*0.85 = 1.37275 from eqn i &ii
moles of C2H6 = 0.313*0.15 = 0.04695
moles of CH4 = 0.676*0.15 = 0.1014
molar composition on wet analysis
% moles of N2 = 0.011 / ( 0.011+1.1067+1.37275+0.04695+0.1014) = 0.011/2.6388= 0.0041
% moles of CO2 = 1.1067 / ( 0.011+1.1067+1.37275+0.04695+0.1014) = 1.1067 / 2.6388 =
0.4193
% moles of H2O = 1.37275 / ( 0.011+1.1067+1.37275+0.04695+0.1014) = 1.37275 / 2.6388=
0.52
% moles of C2H6 = 0.04695 / ( 0.011+1.1067+1.37275+0.04695+0.1014) = 0.04695 /
2.6388=0.0177
% moles of CH4 = 0.1014/ ( 0.011+1.1067+1.37275+0.04695+0.1014)= 0.1014/2.6388 = 0.0384.
2. What factors should be considered in selecting a design strategy.pdf
2. What factors should be considered in selecting a design strategy?
Note: No plagiarism please
Solution
Follwing are the key factors which has to be considered during design strategy.
1. Involvement: Desining a strategy is a process. The main key component in designing a
strategy is involvement of all level of personnel. Involvement of all level of personnel will help
the organization or department to identify extra inputs, insights, challenges, concerns which may
not have taken in account.
2. Communication: During Design Strategy, top down and bottom up approach of
communication should be taken. It starts with bottom up approach by notifying all personnel
about the design strategy. It includes involvement of the personnel in design strategy by means
of group discussion, feedback survey, meeting etc. to identify extra inputs, insights, challenges
which may have unknown. Then this approach is followed by top down approach where top
officials of the organization will look in to the desgn strategy and share this with all the
personnel.
3. Articulation: One of the most Important factor that has to be conidered in Design Strategy is
the articulation. In general, Articulation means clarity of the speech. During communicarion,
there has to articulation in all level of employees. They should be clear about their opinions,
thoughts etc. Any vague statement can lead to a bad design strategy
4. Innovation: Every organization tells their employees to be innovative, creative. But many
organizations fails to develope a strategy for innovation. It is crucial for organization to develope
a strategy for implementing innovations of employees.
5. Management: Once design strategy is fully formed, the next step is the management of the
project. 95% of the employees does not understand the compay\'s strategy. Without
understanding the design strategy, project management is difficult. So high level of involvement
of all employess is crucial for undestanding the design strategy. It increases their sense of
responsibilty and level of commitments to their work..
–PLS write program in c++Recursive Linked List OperationsWrite a.pdf
–PLS write program in c++
Recursive Linked List Operations
Write a C++ to manage a list of string elements. Besides the basic operations expected of a
linked list (add an element, remove an element, check for empty list, report size (number) of
elements in the list, and print out the content of the list), the linked list class also needs to have a
method called sort() and a method called reverse() the manipulate the string elements. The sort
method rearranges the elements in the list so they are sorted in alphabetical order. The reverse
method reverses the order of the elements of the list. The class should use recursion to
implement the sort and reverse operations
Solution
#include
#include
#include
#include
struct node //declaring node
{
int info;
struct node *next;
}*start;
class single_llist //declaring class
{
public:
node* create_node(int);
void insert_begin();
void insert_pos();
void insert_last();
void delete_pos();
void sort();
void reverse();
void display();
single_llist()
{
start = NULL;
}
};
main()
{
int choice, nodes, position, i;
char str[100];
single_llist sl;
start = NULL;
while (1)
{
cout<>choice;
switch(choice)
{
case 1:
cout<<\"Inserting Node at Beginning: \"<info = value;
temp->next = NULL;
return temp;
}
}
/*
* Inserting element in beginning
*/
void single_llist::insert_begin()
{
int value;
cout<<\"Enter the value to be inserted: \";
cin>>value;
struct node *temp, *p;
temp = create_node(value);
if (start == NULL)
{
start = temp;
start->next = NULL;
}
else
{
p = start;
start = temp;
start->next = p;
}
cout<<\"Element Inserted at beginning\"<>value;
struct node *temp, *s;
temp = create_node(value);
s = start;
while (s->next != NULL)
{
s = s->next;
}
temp->next = NULL;
s->next = temp;
cout<<\"Element Inserted at last\"<>value;
struct node *temp, *s, *ptr;
temp = create_node(value);
cout<<\"Enter the postion at which node to be inserted: \";
cin>>pos;
int i;
s = start;
while (s != NULL)
{
s = s->next;
counter++;
}
if (pos == 1)
{
if (start == NULL)
{
start = temp;
start->next = NULL;
}
else
{
ptr = start;
start = temp;
start->next = ptr;
}
}
else if (pos > 1 && pos <= counter)
{
s = start;
for (i = 1; i < pos; i++)
{
ptr = s;
s = s->next;
}
ptr->next = temp;
temp->next = s;
}
else
{
cout<<\"Positon out of range\"<next;s !=NULL;s = s->next)
{
if (strcmp(ptr->info > s->info))
{
value = ptr->info;
ptr->info = s->info;
s->info = value;
}
}
ptr = ptr->next;
}
}
void single_llist::delete_pos() //delete element from given position
{
int pos, i, counter = 0;
if (start == NULL)
{
cout<<\"List is empty\"<>pos;
struct node *s, *ptr;
s = start;
if (pos == 1)
{
start = s->next;
}
else
{
while (s != NULL)
{
s = s->next;
counter++;
}
if (pos > 0 && pos <= counter)
{
s = start;
for (i = 1;i < pos;i++)
{
ptr = s;
s = s->next;
}
ptr->next = s->next;
}
else
{
cout<<\"Position out of range\"<next == NULL)
{
return;
}
ptr1 = start;
ptr2 = ptr1->next;
ptr3 = ptr2->next;
ptr1->next = NULL;
ptr2->next = ptr1;
while (ptr3 != NULL)
{
ptr1 = ptr2;
ptr2.
Wireless networks are considered insecure for many organizations; es.pdf
Wireless networks are considered insecure for many organizations; especially, for the
government agencies. Explain why wireless networks are still considered insecure and what you
would recommend to secure them. Explain your reasons in terms of technologies and policies.
Solution
accidental asscosciation and maliscious assosciation are the two kinds of stratagies that can be
applied on the organization or the individuals to get the work from them as well as benifits to
them. so people should be very carefull when they are using sensible information.
Let us consider the corporate network and if it is attacked by any of them then that occurs from
different perspectives. So it is called as “accidental association”.
“Malicious associations” can be interpreted as like this for example if the wireless devices are
going to be connected to the organization network using laptop by the intruders on behalfof of a
organizations access point (AP).
there are many other threats like
man in the middlea attack
denial of service etc..
caffe latte attack.
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KKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKK
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Read| The latest issue of The Challenger is here! We are thrilled to announce that our school paper has qualified for the NATIONAL SCHOOLS PRESS CONFERENCE (NSPC) 2024. Thank you for your unwavering support and trust. Dive into the stories that made us stand out!
Synthetic Fiber Construction in lab .pptx
Synthetic fiber production is a fascinating and complex field that blends chemistry, engineering, and environmental science. By understanding these aspects, students can gain a comprehensive view of synthetic fiber production, its impact on society and the environment, and the potential for future innovations. Synthetic fibers play a crucial role in modern society, impacting various aspects of daily life, industry, and the environment. ynthetic fibers are integral to modern life, offering a range of benefits from cost-effectiveness and versatility to innovative applications and performance characteristics. While they pose environmental challenges, ongoing research and development aim to create more sustainable and eco-friendly alternatives. Understanding the importance of synthetic fibers helps in appreciating their role in the economy, industry, and daily life, while also emphasizing the need for sustainable practices and innovation.
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Acetabularia Information For Class 9 .docx
Acetabularia acetabulum is a single-celled green alga that in its vegetative state is morphologically differentiated into a basal rhizoid and an axially elongated stalk, which bears whorls of branching hairs. The single diploid nucleus resides in the rhizoid.
A Strategic Approach: GenAI in Education
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
How to Make a Field invisible in Odoo 17
It is possible to hide or invisible some fields in odoo. Commonly using “invisible” attribute in the field definition to invisible the fields. This slide will show how to make a field invisible in odoo 17.
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- 1. Why is mild to moderate expression of an X-linked recessive trait not observed in heterozygous female Drosophila? Solution In X-linked recessive trait, females will get the mutant phenotype if they receive both copies of the defective gene or females who are homozygous for the gene mutation. If they have one dominant and one recessive allele, they will not show any mutant phenotype. SSex-linkedgenes are located on the on the X chromosomes, for example wild type of drosophila will have red eyes, and white color eyes are recessive. If a female drosophila possesses XRXr(heterozygous condition), drosophila eyes will be red in color, if gets both the recessive alleles XrXr(homozygous condition), then only her eyes will become white. That is the reason why heterozygous female drosophila will not express the mild to moderate X-linked recessive trait. Due to the presence of two copies of X chromosomes, only one copy of wild allele is enough for them to get normal phenotype.