Low platelet count it a recessively inherited trait. Reevaluation of .pdf
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Low platelet count it a recessively inherited trait. Reevaluation of the genetic screen indicates that Nikoleta is homozygous dominant normal for a trait that causes low platelet count and her husband is homozygous recessive If the F, offspring has normal platelets and beta-thalassemia. What are the parental gametes? Based on the F_1 progeny phenotype given and the parental gametes, write the two possible genotypes. What is the gamete frequency it the two genotypes in (b) were crossed with each other? What is the phenotype frequency for the cross in (c)? Use the evidence from the case to advice Nikoleta and her husband about having a family. Nikoleta and her husband Andrew conceived four times. What are the possible combinations for a boy and get for each time Mrs. Stamos became pregnant? Solution Legends- XX-female XY-male HH/hh-homozygous trait for platelets count Hh-heterozygous trait for platelets count 1.a. The parental gametes will be – XHXH - homozygous dominant mother XhY- homozygous recessive b. possible genotype are- 2XHXh and 2XHY c. gamete frequency for XHXh will be 2/4=1/2 gamete frequency for XHY will be 2/4=1/2 d. phenotype frequency is as follows Parents Homozygous dominant mother Homozygous recessive father XHXH XhY Gametes XH XH Xh Y Male/female XH XH Xh XHXh heterozygous dominant XHXh heterozygous dominant Y XHY heterozygous dominant XHY heterozygous dominant F1 progeny 2. From the f1 progeny it is clear that all offspring are heterozygous, as platelets count is homozygous recessive trait. All children will be normal. Male/female XH XH Xh XHXh heterozygous dominant XHXh heterozygous dominant Y XHY heterozygous dominant XHY heterozygous dominant.
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4.3 theoretical genetics
This document discusses genetics concepts including:
- Genotypes describe an organism's alleles while phenotypes describe its observable characteristics.
- Dominant alleles will be expressed whether homozygous or heterozygous, while recessive alleles only express when homozygous.
- Codominance occurs when both alleles in a heterozygote affect the phenotype.
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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
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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
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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
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father Is homozygous normal? The infant inherited both copies of chromosome 1 from the
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chromosome 22 generated a recessive phenotype. Plasmid DNA, like bacterial DNA: is circular.
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4.3 theoretical genetics
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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
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the no disjunction that gave rise to the young man with Klinefelter syndrome take place? Assume
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namespace ROCK_PAPER_SCISSORS_GAME_2
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public partial class Form1 : Form
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public Form1()
{
InitializeComponent();
}
sim iComputer As Integer = 0;
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Private iDraw As Integer = 0;
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playerLabel.Text = \"Rock\";
PicBox.Image = \"Rock.bmp\";
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computerLabel.Text = \"Paper\";
scoreLabel.Text = \"You lose!\";
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}
if (iComputer == 2)
{
pic.Image = \"Rock.bmp\";
computerLabel.Text = \"Rock\";
scoreLabel.Text = \"Draw!\";
iDraw = iDraw + 1;
}
if (iComputer = 3)
{
.Image = \"Scissors.bmp\";
computerLabel.Text = \"Scissors\";
scoreLabel.Text = \"You win!\";
iWins = iWins + 1;
}
{
LabelWins1.Text = iWins;
LabelDraw1.Text = iDraw;
LabelLose1.Text = iLose;
}
}
private void paperButton_Click(object sender, EventArgs e)
{
{
playerLabel.Text = \"Paper\";
int iComputer = rand.Next(3);
computerLabel.Text = iComputer;
}
if (iComputer == 1)
{
computerLabel.Text = \"Paper\";
scoreLabel.Text = \"Draw\";
iDraw = iDraw + 1;
}
if (iComputer == 2)
{
computerLabel.Text = \"Paper\";
scoreLabel.Text = \"Draw\";
iDraw = iDraw + 1;
}
if (iComputer == 3)
{
computerLabel.Text = \"Scissor\";
scoreLabel.Text = \"You lose!\";
iLose.Text = iLose + 1;
}
{
LabelWins1.Text = iWins;
LabelDraw1.Text = iDraw;
LabelLose1.Text = iLose;
}
}
private void scissorsButton_Click(object sender, EventArgs e)
{
{
int iComputer = rand.Next(3);
computerLabel.Text = iComputer;
LabelWins1.Text = iWins;
LabelDraw1.Text = iDraw;
LabelLose.Text = iLose;
}
if (iComputer == 1)
{
computerLabel.Text = \"Paper\";
scoreLabel.Text = \"You win!\";
iWins = iWins + 1;
}
if (iComputer == 2)
{
computerLabel.Text = \"Rock\";
scoreLabel.Text = \"You lose!\";
iLose = iLose + 1;
}
if (iComputer == 3)
{
computerLabel.Text = \"Scissor\";
scoreLabel.Text = \"Draw\";
iDraw = iDraw + 1;
}
{
LabelWins1.Text = iWins;
LabelDraw1.Text = iDraw;
LabelLose1.Text = iLose;
}
}
private void resetButton_Click(object sender, EventArgs e)
{
iLose = 0;
iDraw = 0;
iWins = 0;
LabelWins1.Text = iWins;
LabelDraw1.Text = iDraw;
LabelLose1.Text = iLose;
playerLabel.Text = \"\";
computerLabel.Text = \"\";
picComputer.Image = Nothing;
picPlayer.Image = Nothing;
scoreLabel.Text = \"\";
}
}
private void DisplayImages(int randomImage, PictureBox picBox)
// Get the current Application working directory.
string imgPath = Environment.CurrentDirectory
+ \"\\\\picBoxImage\\\\\";
// Use switch case to display the respective images in PictureBox.
switch (randomImage)
{
case 1:
// picBox.ImageLocation =
// imgPath + \"Rock.BMP\";
// break;
picBox.Image = ROCK_PAPER_SCISSORS_GAME_2.
Properties.Resources.Rock;
break;
case 2:
// picBox.ImageLocation =
// imgPath + \"Paper.BMP\";
// br.
Describe tolerance as a immunologic function. What is the consequenc.pdf
Describe tolerance as a immunologic function. What is the consequence of loss of tolerance? ( ¾
page? 2 solid paragraphs)
Solution
Immune tolerance is an act of self tolerance. Self tolerance could be seen against certain
antigens. Tolerance is a physiological mechanism. Due to immune tolerance, the body response
to an antigen either decreases or completely get eliminated. Due to this self tolerance, it leads to
suppressing of allergic responses against antigens. Due to this self tolerance, the host in which
the immune tolerance observed shows chronic infections instead of their elimination. Due to this
self tolerance within the host, the auto reactive lymphocytes become inactivated. Within the
mother due to immune-tolerance, the attack on fetuses has been severe. Due to this imbalance,
leads to autoimmunity through triggering T-helper cell functioning. When the body loses its
tolerance to self antigens, leads to autoimmune disorders such as thyroiditis, multiple sclerosis,
rheumatoid arthritis and systemic lupus erythematosus..
More Related Content
Similar to Low platelet count it a recessively inherited trait. Reevaluation of .pdf
Biol102 chp15-pp-spr10-100412104754-phpapp01
This document summarizes key concepts about the chromosomal basis of inheritance from genetics. It discusses how genes are located on chromosomes and segregate during meiosis according to Mendel's laws. It describes classic experiments by Morgan using fruit flies that provided evidence linking genes to chromosomes. The document explains genetic linkage, recombination, and how crossover during meiosis generates genetic variation. It also discusses different types of chromosomal abnormalities and how they can cause genetic disorders.
Biol102 chp15-pp-spr10-100412104754-phpapp01
This document summarizes key concepts about the chromosomal basis of inheritance from genetics research in the early 20th century. It discusses how Mendel's laws of inheritance can be explained by the behavior of chromosomes and genes during meiosis and fertilization. The work of Thomas Hunt Morgan with fruit flies provided evidence that genes are located on chromosomes. His experiments showed that linked genes tend to be inherited together but can assort independently at low frequencies, due to chromosome crossover during meiosis. This led to the theory that genes reside at specific loci on chromosomes and explained how genetic variation arises.
Sex Linked Inheritance in Man
This document discusses sex-linked inheritance in humans. It begins by introducing sexual reproduction and chromosomes. Males have XY sex chromosomes and are heterogametic, while females have XX and are homogametic. Autosomes carry genes for somatic traits, while sex chromosomes determine sex and carry some somatic genes. These sex-linked genes show patterns of inheritance different from autosomal genes. Examples given are color blindness and hemophilia, which are recessive traits linked to the X chromosome. Diagrams show possible genotypes and phenotypes that can result when parents with normal, affected, or carrier genotypes reproduce. The conclusion emphasizes that case studies show possible outcomes, not fixed probabilities, for offspring.
Presentation 02 genetics i
The document discusses several of Gregor Mendel's discoveries and laws of genetics from his experiments with pea plants including:
1) Mendel's law of segregation which states that alleles for a gene separate into gametes during reproduction.
2) His law of independent assortment which states that different genes assort independently if located on separate chromosomes.
3) That dominant alleles are fully expressed in heterozygotes while recessive alleles have no visible effect.
1. In a given plant, the yellow seed color allele (Y) is dominant to .pdf
1. In a given plant, the yellow seed color allele (Y) is dominant to green seed color (y). Give the
genotypic and phenotypic ratios for the following crosses? a. homozygous yellow x green b.
heterozygous yellow x green c. heterozygous yellow x heterozygous yellow d. green x green 2. If
two animals heterozygous for a single pair of genes are mated and have 276 offspring, how many
would be expected to have the dominant genotype? 3. Two long-winged fies are mated. The
offspring included 90 with long wings and 32 with short wings. Which phenotype is recessive?
4. A blood test reveals that a couple both have the sickle-cell trat. What is the probability of their
producing a child with sickle-cell anemia? 5. A blood type A woman has a type B child. She
accuses a type O man of being the father. Knowing genetics you are asked to give an opinion; do
so. 6. Assume that human eye color is controlled by a single pair of alleles. A blue-eyed man,
both of whose parents were bom brown-eyed, marries a brown-eyed woman whose father was
blue-eyed and whose mother was brown-eyed. They have a blue-eyed child. Which are the
genotypes of all individuals mentioned? 7. A red fruit, tall tomato plant RrTt, is crossed with a
yellow-fruited dwarf plant ritt. What genotypes and phenotypes would be expected in the
offispring? 8. What genotype and phenotype would be expected in the offspring if the cross in
the previous problem had been RrTt x RrTt ?.
Genetics practice q key
- One or more of the three children will have disease 1.
- All three children will have the disease.
- At least one child out of three will be phenotypically normal.
Linked inheritance
Breaking down Biology into simpler bits is the most effective way to learn hence this presentation aims to simplify the concept of 'Linked Inheritance' which makes understanding Inheritance better.
Please give an explaination fir those quesions! 2 3 4 5 1- Aman with r.docx
Please give an explaination fir those quesions!
2
3
4
5
1. Aman with red.geen colour blindness is marned to a girl with normal colwir vision. What is the probaliliny that his childron will be coluur blind? 2. A boy has Dachenne muscular dyatrophy. His malemal uncle had died of museular dystrophy at the age of 21 years. What is the prohatility that this chilfs sibs will be affected? Assignments \#3 rocosive or S -kinked dominant and X -linked recesive inheritance. (Nole that a potigrte may be coempable with muve thas ane type of inheritance.) Assignnents \#4 Examine the pedieree frum a funily with a ernetic disease and anvv the questiens below: a. Does this pedigree isdicate autosonal deuninant, teoessive of set-link type of inheritance? Give reasons for your choice. b. Assuaning that B and b are the normal and mataut alleies respectively what would be the enenotypes of the individuals :II.1, 11 2 and III .3? c. Individual II. 3 requested genctic counselling. What is the probubility that her child woald be affected. Fxplain why. Assignments \#5 Susan's grandfather was deaf, and passed down a hercditary form of deafhess within Susan's family as shown in Figure. A. Is this mutation most likely to be dominant of recessive?? B. Is it carried on an autosome of a sex cluomosome? Why? C. A complete SNP analysis has been done for all of the 11 grandchildren (4 affected, and 7 unaffected) In comparing these 11 SNP result, how long a haplotype block would you expect to find around the critical \#ene? How might you deleet it? Assignments \#1 Distinguish between the following genetic terms: A. Gene and allele. B. Homozygous and heterozygous. C. Genotype and phenotype. D. Dominant and recessive. 1. A man with red-green colour blindness is married to a girl with normal colour vision. What is the probability that his children will be colour blind? 2. A boy has Duchenne muscular dystrophy. His maternal uncle had died of muscular dystrophy at the age of 21 years. What is the probability that this child's sibs will be affected? Explain, giving reasons, whether the following pedigrees are compatible with autosomal dominant, autosomal recessive or X -linked dominant and X -linked recessive inheritance. (Note that a pedigree may be compatible with more than one type of inheritance.) a. b. c. a. b. c. Assignments \#4 Examine the pedigree from a family with a genetic disease and ansv the questions below: a. Does this pedigree indicate autosomal dominant, recessive or sex-link type of inheritance? Give reasons for your choice. b. Assuming that B and b are the normal and mutant alleles respectively what would be the genotypes of the individuals: II.I, II.2 and III.3 ? c. Individual II. 3 requested genetic counselling. What is the probability that her child would be affected. Explain why. Susan's grandfather was deaf, and passed down a hereditary form of deafness within Susan's family as shown in Figure. A. Is this mutation most likely to be dominant or recessive? B. I.
Sex determination and nondisjunction
Sex determination in humans and many other mammals is genetic, with the presence or absence of the Y chromosome determining maleness or femaleness. The SRY gene on the Y chromosome causes embryonic gonads to develop into testes rather than ovaries. In birds and some fish, the ZW sex determination system is used instead of XY, with females being ZW and males ZZ. Genes located on the sex chromosomes are called sex-linked genes and can be inherited in sex-specific patterns. Nondisjunction of sex chromosomes during meiosis can sometimes result in individuals with abnormal sex chromosome complements.
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.
Module 10 human heredity
- Karyotypes show the 23 pairs of chromosomes in human cells, with autosomes making up the first 22 pairs and sex chromosomes determining gender as the 23rd pair (XX female, XY male)
- Pedigree charts show the genetic history and inheritance of traits in a family over generations using symbols to represent individuals and their genotypes
- Traits can be dominant, recessive, or sex-linked (located on the X chromosome); the chance of offspring inheriting traits can be determined using Punnett squares
- Down syndrome is identified via karyotype as having an extra copy of chromosome 21, giving 47 total chromosomes rather than the typical 46
Module 10 human heredity
- Karyotypes show the 23 pairs of chromosomes in human cells, with autosomes making up the first 22 pairs and sex chromosomes determining gender as the 23rd pair (XX female, XY male)
- Pedigree charts show the genetic history and inheritance of traits in a family over generations using symbols to represent individuals and their genotypes
- Traits can be dominant, recessive, or sex-linked (located on the X chromosome); the chance of offspring inheriting traits can be determined using Punnett squares
- Down syndrome is identified via karyotype as having an extra copy of chromosome 21, giving 47 total chromosomes rather than the typical 46
Module 10 human heredity
- Karyotypes show the 23 pairs of chromosomes in human cells, with autosomes making up the first 22 pairs and sex chromosomes determining gender as the 23rd pair (XX female, XY male)
- Pedigree charts show the genetic history and inheritance of traits in a family over generations using symbols to represent individuals and their genotypes
- Traits can be dominant, recessive, or sex-linked (located on the X chromosome); the chance of offspring inheriting traits can be determined using Punnett squares
- Down syndrome is identified via karyotype as having an extra copy of chromosome 21, giving 47 total chromosomes rather than the typical 46
Please- answers all my question is for genetic class 3- Describe the.docx
Please, answers all my question is for genetic class
3. Describe the pattern of inheritance shown in the pedigree below, giving evidence to support your answer. Give the genotypes for individuals II-2, II-4 and II-2. Include a legend for the symbols used. (10 marks) Describe the pattern of inheritance shown in the pedigree below, giving evidence to support your answer. Give the genotypes for individuals II-4, II-7 and III-7. Include a legend for the symbols used. ( 10 marks) 5. A type of hemophilia is due to an X-linked recessive allele. An affected man whose phenotype has been successfully managed now has children with a woman who is a carrier. Providing a legend for your, answer the following questions: a. What are the genotypes of the mother and father? (1 mark) b. What are the probabilities that i. a child will be affected with hemophilia. ii. a child is a carrier. iii. their daughter is a carrier. iv. their male fetus will be affected. (4 marks) 6. Genes for two different recessive disorders are linked on the X chromosome; they are 26 map units apart. A woman is a carrier for both disorder A and disorder B , with the alleles for these genes in cis. Her husband has normal alleles for both of these genes. The woman is pregnant with their first son. What is the probability (in percent) this son is affected with both disorders? Explain, showing the proportions and types of the woman's gametes and identifying them as parental or recombinant. (5 marks) 7. In certain human matings of albino and black-haired parents, the offspring are always blackhaired. When these black-haired offspring of the different families mate with each other (non-consanguineously) and their children are pooled to get significant data, these are the children's phenotypic proportions: 9 black-haired children 3 brown-haired children 4 albino children a. How many genes are involved in this trait? b. How many alleles of each gene are there? c. What type of gene interaction is occurring for this trait? Describe it. d. What are the genotypes of the initial albino and black-haired parents? Provide a legend. e. What are the genotypes of their offspring (all black-haired)? f. What are the genotypes and the specific proportions of the albino children in the third generation? g. What are the genotypes of the brown-haired children? 8. Which type of chromosomal aneuploidy typically has less severe effects on an individual: sex chromosome aneuploidy or autosomal aneuploidy? Explain why. (4 marks) 9. An organism has six chromosomes in its somatic cells. Determine the number of gametes with different chromosome compositions that this organism can potentially create. Explain your answer, including the genetic basis for your calculation. Ignore the allelic variation created by crossing over. ( 4 marks) 10. Compare the inheritance and transmission of Xlinked genes with that of mitochondrial genes. (4 marks) 11. A woman has a normal number of fingers and toes. She has many relatives with ex.
Genetics Notes
Genetics is the study of genes and heredity. The document discusses key genetics concepts including chromosomes, DNA, genes, alleles, dominant and recessive traits, Punnett squares, sex-linked inheritance, and pedigrees. It provides examples to illustrate genetic inheritance patterns and the use of tools like Punnett squares and pedigrees to predict offspring traits.
Gamified Intro to Genetics
This document contains genetics problems involving monohybrid and dihybrid crosses. It begins with background on genes, alleles, phenotypes, and Mendel's laws of inheritance. It then presents 5 genetics problems involving traits such as body color, head shape, nose length, blood type, and baldness. For each problem, it asks for the genotypes and phenotypes of offspring from a cross, as well as the phenotypic and genotypic ratios in subsequent generations. It concludes by asking for the importance of genetics and an explanation of one of Mendel's laws of inheritance.
Pedigree problems (NEET)
The document describes a pedigree analysis to determine the mode of inheritance shown. It analyzes a pedigree where normal parents produce affected offspring in generations II and III. This indicates recessive inheritance. As it skips generations and not all offspring are affected, it also points to recessive inheritance. Autosomal recessive and X-linked recessive are considered, but X-linked recessive is ruled out. Therefore, the mode of inheritance is determined to be autosomal recessive.
Tatiana and David workshop
The document summarizes several genetics problems involving monohybrid and dihybrid crosses between organisms with different traits.
For a monohybrid cross between a homozygous red-eyed rat and a homozygous black-eyed rat, all offspring would have black eyes.
A dihybrid cross between a heterozygous long-winged, big-peaked bird and a homozygous short-winged, small-peaked bird would produce offspring that are 25% each of long-winged/big peak, long-winged/small peak, short-winged/big peak, and short-winged/small peak.
A cross involving a male with sex-linked
Evolution lectures WK9
The document discusses genetic drift and inbreeding in the Amish population. It provides three key points:
1) The Amish population originated from around 400 founders and has since expanded, making it a valuable population for genetic studies due to excellent records and a restricted gene pool. This small founder population size has led to significant genetic drift.
2) Several disadvantageous alleles have drifted to high frequencies in the Amish despite being selected against, demonstrating that genetic drift affects all loci.
3) Inbreeding is defined as non-random mating between relatives, increasing identity by descent; however, the Amish avoid close cousin marriages, resulting in less inbreeding than expected from random mating.
Basics of Genetics presented by waseem javid
This document provides an overview of genetics concepts including:
- Gregor Mendel is considered the "Father of Genetics" and discovered the principles of inheritance including independent assortment.
- Traits are determined by genes located on chromosomes, which are received in homologous pairs from each parent. Genes can be dominant, recessive, or codominant.
- A Punnett square can be used to predict the possible genotypes and phenotypes of offspring from a genetic cross based on the parents' genotypes.
- Sex is determined by sex chromosomes, with females typically having XX and males having XY. Sex-linked traits are located on the X chromosome.
- Pedigrees can represent the inheritance of traits through generations
Similar to Low platelet count it a recessively inherited trait. Reevaluation of .pdf (20)
1. In a given plant, the yellow seed color allele (Y) is dominant to .pdf
1. In a given plant, the yellow seed color allele (Y) is dominant to .pdf
Please give an explaination fir those quesions! 2 3 4 5 1- Aman with r.docx
Please give an explaination fir those quesions! 2 3 4 5 1- Aman with r.docx
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More from fasttracksunglass
I have been working on this ROCK, PAPER, SCISSORS project for the pa.pdf
I have been working on this ROCK, PAPER, SCISSORS project for the past 6 days or so, and
it\'s still giving me absolute fits. I have created two pictureBox controls in the designer, and here
is my current source
code:
namespace ROCK_PAPER_SCISSORS_GAME_2
{
public partial class Form1 : Form
{
public Form1()
{
InitializeComponent();
}
sim iComputer As Integer = 0;
Private iWins As Integer = 0;
Private iDraw As Integer = 0;
Private iLose As Integer = 0;
private void rockButton_Click(object sender, EventArgs e)
{
{
playerLabel.Text = \"Rock\";
PicBox.Image = \"Rock.bmp\";
int iComputer = rand.Next(3);
computerLabel.Text = \"Paper\";
scoreLabel.Text = \"You lose!\";
iLose = iLose + 1;
}
if (iComputer == 2)
{
pic.Image = \"Rock.bmp\";
computerLabel.Text = \"Rock\";
scoreLabel.Text = \"Draw!\";
iDraw = iDraw + 1;
}
if (iComputer = 3)
{
.Image = \"Scissors.bmp\";
computerLabel.Text = \"Scissors\";
scoreLabel.Text = \"You win!\";
iWins = iWins + 1;
}
{
LabelWins1.Text = iWins;
LabelDraw1.Text = iDraw;
LabelLose1.Text = iLose;
}
}
private void paperButton_Click(object sender, EventArgs e)
{
{
playerLabel.Text = \"Paper\";
int iComputer = rand.Next(3);
computerLabel.Text = iComputer;
}
if (iComputer == 1)
{
computerLabel.Text = \"Paper\";
scoreLabel.Text = \"Draw\";
iDraw = iDraw + 1;
}
if (iComputer == 2)
{
computerLabel.Text = \"Paper\";
scoreLabel.Text = \"Draw\";
iDraw = iDraw + 1;
}
if (iComputer == 3)
{
computerLabel.Text = \"Scissor\";
scoreLabel.Text = \"You lose!\";
iLose.Text = iLose + 1;
}
{
LabelWins1.Text = iWins;
LabelDraw1.Text = iDraw;
LabelLose1.Text = iLose;
}
}
private void scissorsButton_Click(object sender, EventArgs e)
{
{
int iComputer = rand.Next(3);
computerLabel.Text = iComputer;
LabelWins1.Text = iWins;
LabelDraw1.Text = iDraw;
LabelLose.Text = iLose;
}
if (iComputer == 1)
{
computerLabel.Text = \"Paper\";
scoreLabel.Text = \"You win!\";
iWins = iWins + 1;
}
if (iComputer == 2)
{
computerLabel.Text = \"Rock\";
scoreLabel.Text = \"You lose!\";
iLose = iLose + 1;
}
if (iComputer == 3)
{
computerLabel.Text = \"Scissor\";
scoreLabel.Text = \"Draw\";
iDraw = iDraw + 1;
}
{
LabelWins1.Text = iWins;
LabelDraw1.Text = iDraw;
LabelLose1.Text = iLose;
}
}
private void resetButton_Click(object sender, EventArgs e)
{
iLose = 0;
iDraw = 0;
iWins = 0;
LabelWins1.Text = iWins;
LabelDraw1.Text = iDraw;
LabelLose1.Text = iLose;
playerLabel.Text = \"\";
computerLabel.Text = \"\";
picComputer.Image = Nothing;
picPlayer.Image = Nothing;
scoreLabel.Text = \"\";
}
}
private void DisplayImages(int randomImage, PictureBox picBox)
// Get the current Application working directory.
string imgPath = Environment.CurrentDirectory
+ \"\\\\picBoxImage\\\\\";
// Use switch case to display the respective images in PictureBox.
switch (randomImage)
{
case 1:
// picBox.ImageLocation =
// imgPath + \"Rock.BMP\";
// break;
picBox.Image = ROCK_PAPER_SCISSORS_GAME_2.
Properties.Resources.Rock;
break;
case 2:
// picBox.ImageLocation =
// imgPath + \"Paper.BMP\";
// br.
Describe tolerance as a immunologic function. What is the consequenc.pdf
Describe tolerance as a immunologic function. What is the consequence of loss of tolerance? ( ¾
page? 2 solid paragraphs)
Solution
Immune tolerance is an act of self tolerance. Self tolerance could be seen against certain
antigens. Tolerance is a physiological mechanism. Due to immune tolerance, the body response
to an antigen either decreases or completely get eliminated. Due to this self tolerance, it leads to
suppressing of allergic responses against antigens. Due to this self tolerance, the host in which
the immune tolerance observed shows chronic infections instead of their elimination. Due to this
self tolerance within the host, the auto reactive lymphocytes become inactivated. Within the
mother due to immune-tolerance, the attack on fetuses has been severe. Due to this imbalance,
leads to autoimmunity through triggering T-helper cell functioning. When the body loses its
tolerance to self antigens, leads to autoimmune disorders such as thyroiditis, multiple sclerosis,
rheumatoid arthritis and systemic lupus erythematosus..
Answer the following question about human evolution as inferred from.pdf
Answer the following question about human evolution as inferred from genome sequencing.
Summarize the demographc history of humans that explains how certain populations acquired
Neandertal and Denisovan DNA. (4 sentence maximum)
Solution
There is a hypothesis that anatomically modern human inbreed with the Neandertal population.
Neandertal genome project presented genetic evidence that interbreeding did took place. Portion
of Neanderthal admixture is present in the DNA of modern Eurasians and Oceanians, and nearly
absent in sub-Saharan African populations. Some part of the genome of Melanesians is derived
from Denisovan DNA. It is possily because of early migration of the ancestors of melanesians
into southeast asia..
40.Classification of a network as a LAN or a WAN is not relevant to .pdf
40.Classification of a network as a LAN or a WAN is not relevant to wireless communication
topologies.
1) True
2) False
44. Which of the following is NOT an example of a business term category that should be
addressed in the contract negotiations?
1) License term, renewal, termination
2) Risk-sharing terms
3) Upgrades and enhancements
4) Interfaces and conversions
45. Which of the following computing attributes is essential to conducting analytical research on
the scope of mapping the human genome?
1) High-speed computation capability
2) Robust data encryption
3) Wireless data transmission
4) Expert system interface
50.The market for financial applications software is BEST described as
1) entrepreneurial.
2) niche.
3) concentrated.
4) volatile.
52.Which of the following describes cellular digital packet data?
1) WAN architecture
2) Infrared technology
3) Used in Internet communication
4) Transmitted on a T1 line
57.What is network physical topology?
1) Configuration of computer and peripheral devices
2) Data transmission protocols
3) Network operating systems
4) Placement of router devices
68.Which of the following describes computer-assisted medical instrumentation?
1) Is used for both data capture and data display
2) Has limited application beyond routine monitoring, such as monitoring of vital signs
3) Is a passive technology used primarily in the clinical laboratory
4) Requires hard-wired configuration
71.Which of the following is an example of a utility program?
1) Virus scan software
2) Multiplexer
3) USB port
4) Digital voice translator
Solution
Please follow the data and description :
40) Wireless Network Topology :
It is a logical topology and shows how the computers connect and interact each other when there
is no physical connection, no cables connecting the computers. The computers communicate
each other directly, using the wireless devices. Wireless networks can have infrastructure or ad
hoc topology. So this does not deals with the types of communication networks that is either a
LAN or a WAN.
So the answer is OPTION 2 (False).
44) Contract negotiation :
It is the process of give and take policy that the parties go through to reach an agreement. Or, in
general as they often say in the terms of business, \"you don\'t get what you deserve; you get
what you negotiate.\" So this does involve with the process of risk management, license terms
and the renewal or the termination of it, any changes during the implementation or evne in the
execution process. But this does not go with the process of interfaces and conversion process.
So the answer is OPTION 4 (Interfaces and COnversions).
45) In the process of computing the mapping for the human genome there are some attributers
that needs to be fulfilled during the process so as to achieve the efficient resutls. Some of them
are the highly relaible data, fast computing, data mining, data warehousing and so on.
So the answer is OPTION 1 (High-speed computation capability).
50) Financial a.
12.9 Program Online shopping cart (continued) (C++)This program e.pdf
12.9 Program: Online shopping cart (continued) (C++)
This program extends the earlier \"Online shopping cart\" program. (Consider first saving your
earlier program).
(1) Extend the ItemToPurchase class per the following specifications:
Parameterized constructor to assign item name, item description, item price, and item quantity
(default values of 0). (1 pt)
Public member functions
SetDescription() mutator & GetDescription() accessor (2 pts)
PrintItemCost() - Outputs the item name followed by the quantity, price, and subtotal
PrintItemDescription() - Outputs the item name and description
Private data members
string itemDescription - Initialized in default constructor to \"none\"
Ex. of PrintItemCost() output:
Ex. of PrintItemDescription() output:
(2) Create three new files:
ShoppingCart.h - Class declaration
ShoppingCart.cpp - Class definition
main.cpp - main() function (Note: main()\'s functionality differs from the warm up)
Build the ShoppingCart class with the following specifications. Note: Some can be function
stubs (empty functions) initially, to be completed in later steps.
Default constructor
Parameterized constructor which takes the customer name and date as parameters (1 pt)
Private data members
string customerName - Initialized in default constructor to \"none\"
string currentDate - Initialized in default constructor to \"January 1, 2016\"
vector < ItemToPurchase > cartItems
Public member functions
GetCustomerName() accessor (1 pt)
GetDate() accessor (1 pt)
AddItem()
Adds an item to cartItems vector. Has parameter ItemToPurchase. Does not return anything.
RemoveItem()
Removes item from cartItems vector. Has a string (an item\'s name) parameter. Does not return
anything.
If item name cannot be found, output this message: Item not found in cart. Nothing removed.
ModifyItem()
Modifies an item\'s description, price, and/or quantity. Has parameter ItemToPurchase. Does not
return anything.
If item can be found (by name) in cart, check if parameter has default values for description,
price, and quantity. If not, modify item in cart.
If item cannot be found (by name) in cart, output this message: Item not found in cart. Nothing
modified.
GetNumItemsInCart() (2 pts)
Returns quantity of all items in cart. Has no parameters.
GetCostOfCart() (2 pts)
Determines and returns the total cost of items in cart. Has no parameters.
PrintTotal()
Outputs total of objects in cart.
If cart is empty, output this message: SHOPPING CART IS EMPTY
PrintDescriptions()
Outputs each item\'s description.
Ex. of PrintTotal() output:
Ex. of PrintDescriptions() output:
(3) In main(), prompt the user for a customer\'s name and today\'s date. Output the name and
date. Create an object of type ShoppingCart. (1 pt)
Ex.
(4) Implement the PrintMenu() function. PrintMenu() has a ShoppingCart parameter, and
outputs a menu of options to manipulate the shopping cart. Each option is represented by a single
character. Build and output the menu within the function.
If the an inval.
A sterile item is free ofmicrobes.endospores.viruses.prions..pdf
A sterile item is free of
microbes.
endospores.
viruses.
prions.
microbes, endospores, AND viruses.
Solution
Answer :- option (microbes, endospores, viruses.
Explanation :- The sterile item means free from microorganisms like microbes and viruses or
endoapores. Sterilisation is done through radiations or autoclaving..
Both mitochondria and chloroplasts... A. obtain electrons from water .pdf
Both mitochondria and chloroplasts... A. obtain electrons from water B. are surrounded by a
single membrane C. release oxygen as a by-product D. use chemiosmosis to produce ATP E. use
a hydrogen ion concentration gradient to produce ATP F. reduce NAD+, forming NADH Hide
response Change response D
Solution
1. Answer is D.
Both mitochondria and chloroplast use chemiosmosis to produce ATP.
Chemiosmosis is the movement of ions across the semi permeable membrane down their
electrochemical gradient.
Chemiosmosis can be seen in mitochondria and chloroplast and as well as in most bacteria and
archaea.
Mitochondria use the energy of NADH and FADH2 while the chloroplast uses theNADPH.
Protons are produced when the splitting of light occurs that gives a proton gradient.
2. Answer is A.
As the electrons flow along the electron transport chain during cellular respiration the pH of
intermediate space decreases.
In mitochondria , the flow of electrons from NADH and FADH2 to oxygen molecule.
It is coupled to the uphill transport of protons from the matrix across the inner membrane of
mitochondria to the intermembrane space giving the proto motive force..
Please answer all parts thank you!Parents PrPr YY rr x PrPr.pdf
Please answer all parts thank you!
Parents: Pr/Pr Y/Y r/r x Pr/Pr y/y R/R
What are the phenotypes of the parents?_______________________________________
F1 genotype:___________________________F1 phenotype:______________________
Draw a Punnett Square representing the F2 generation:
Parents: Y/Y Pr/Pr R/R x Y/Y pr/pr r/r
What are the phenotypes of the parents?_______________________________________
F1 genotype:___________________________F1 phenotype:______________________
Draw a Punnett Square representing the F2 generation:
Solution
DominantRecessive
Purple (Pr) Red (pr)
Yellow (Y) White (y)
Coloured (R) Colourless (r)
1. Phenotypes of the parents:
Pr/Pr Y/Y r/r X Pr/Pr y/y R/R
purple, yellow, colourless X purple, white, coloured
F1 genotype : Pr Y r X Pr y R
F1 phenotype : purple yellow colourless X purple white coloured
F2 : ----------R ----------> Pr Y R (Purple, Yellow, Coloured)
--------- Y
----------- r -----------> Pr Y r (Purple yellow colourless)
Pr-------
----------R ----------> Pr y R (purple white coloured)
-------- y
----------- r -----------> Pr y r (purple white colourless)
2. Phenotypes of the parents:
Y/Y Pr/Pr R/R x Y/Y pr/pr r/r
yellow, purple, coloured X yellow, red, colourless
F1 genotype : Y Pr R X Y pr r
F1 phenotype : yellow, purple, coloured X yellow, red, colourless
F2 : ----------R ----------> Y Pr R (Yellow, Purple, Coloured)
-------- Pr
----------- r -----------> Y Pr r ( yellow Purple colourless)
Y-------
----------R ---------->Y pr R (yellow red coloured)
-------- pr
--------- r -----------> Y pr r (yellow red colourless)
Apart from punnett square we can also know the F2 gametes by this above method. since I am
not having a scaner right now to represent this problem through punnett square method, I am
representing it like this. sorry for the inconvience if any..
Lists can contain other lists as elements. For example the list HAIR.pdf
The document describes a "multilevel" list called HAIRY that contains other lists as elements. It presents a problem to write a function called flatten that takes a possibly multi-level list as an argument and returns a new single-level list containing the same elements in the same overall order. The function is demonstrated by flattening HAIRY into a single list of its elements.
Match the modified stem to its correct definition.StolonRhizome.pdf
Match the modified stem to its correct definition.
Stolon
Rhizome
Tuber
Bulb
Corm
Stem succulence
Cladophyll
A simple modified stem that stores starch below ground, like a potato.
A belowground modified stem consisting of many compressed internodes with whorls of leaves
surrounding and protecting the stem.
Stems that have become modified to store water and serve as a primary photosynthetic organ in
desert environments.
A stem that has been modified to resemble a leaf.
A modified stem that grows horizontally below the surface of the ground.
A modified stem that grows horizontally above the surface of the ground.
A belowground modified stem for storing starch consisting of a single thick internode wrapped
in papery leaves. - A. B. C. D. E. F. G.
Stolon - A. B. C. D. E. F. G.
Rhizome - A. B. C. D. E. F. G.
Tuber - A. B. C. D. E. F. G.
Bulb - A. B. C. D. E. F. G.
Corm - A. B. C. D. E. F. G.
Stem succulence - A. B. C. D. E. F. G.
CladophyllA.
A simple modified stem that stores starch below ground, like a potato.B.
A belowground modified stem consisting of many compressed internodes with whorls of leaves
surrounding and protecting the stem.C.
Stems that have become modified to store water and serve as a primary photosynthetic organ in
desert environments.D.
A stem that has been modified to resemble a leaf.E.
A modified stem that grows horizontally below the surface of the ground.F.
A modified stem that grows horizontally above the surface of the ground.G.
A belowground modified stem for storing starch consisting of a single thick internode wrapped
in papery leaves.
Solution
Stem succulence - a modified stem to store water and serve as a primary photosynthesis organ in
desert environment and condition..!!
Cladophyll - I is the stem which is being modified to resemble as a leaf which can perform
photosynthesis too..!! Eg:- Asparagus.
Bulb- a below ground modified stem consisting of many compressed internode with whorls of
leaves surrounding and protecting the main interior middle stem.. Eg:- onion..!!
Tuber- a simple modified stem that stores starch below ground, like a potato..!!
Corm- a below ground modified stem for storing starch consisting of a single thick internode
wrapped in papery leaves..!!
Rhizome - a modified stem that grows horizontally below the surface of the ground which can
form adventitious roots and shoots..
Stolon - a modified stem that grows horizontal above the surface of the ground.. They are called
runner...
JAVAAdd to the code at the bottom to do the following two things.pdf
JAVA:
Add to the code at the bottom to do the following two things
1. A \"Monitoring.java\" class that contains a “main(…)” method that will instantiate a
“animalsHabitat” object and call the methods in it.
2. Proper comments and formatting in this .java file and the one that you will create.
PROMPT:
Create a program for a zookeeper monitoring system using two classes. One to be used to
determine what the user wants to do, monitor animals or habitats. The other to read the file for
the chosen action (listed below as animal/habitat file). Create a monitoring system that does all
of the following:
1) Asks a user if they want to monitor an animal, monitor a habitat, or exit
2)Displays a list of animal/habitat options (based on the previous selection) as read from either
the animals or habitats file o Asks the user to enter one of the options
3) Displays the monitoring information by finding the appropriate section in the file
4) Separates sections by the category and selection (such as “Animal - Lion” or “Habitat -
Penguin”)
5) Uses a dialog box to alert the zookeeper if the monitor detects something out of the normal
range (These will be denoted in the files by a new line starting with *****. Do not display the
asterisks in the dialog.)
6) Allows a user to return to the original options
Animals file: (saved on local)
Details on lions
Details on tigers
Details on bears
Details on giraffes
Animal - Lion
Name: Leo
Age: 5
*****Health concerns: Cut on left front paw
Feeding schedule: Twice daily
Animal - Tiger
Name: Maj
Age: 15
Health concerns: None
Feeding schedule: 3x daily
Animal - Bear
Name: Baloo
Age: 1
Health concerns: None
*****Feeding schedule: None on record
Animal - Giraffe
Name: Spots
Age: 12
Health concerns: None
Feeding schedule: Grazing
Habitat file: (Saved on local)
Details on penguin habitat
Details on bird house
Details on aquarium
Habitat - Penguin
Temperature: Freezing
*****Food source: Fish in water running low
Cleanliness: Passed
Habitat - Bird
Temperature: Moderate
Food source: Natural from environment
Cleanliness: Passed
Habitat - Aquarium
Temperature: Varies with output temperature
Food source: Added daily
*****Cleanliness: Needs cleaning from algae
Code so far:
package monitoring;
import java.io.FileInputStream;
import java.io.IOException;
import java.util.*;
public class animalsHabitat {
private String filePath;
final private Scanner scnr;
public animalsHabitat() {
filePath = \"C:\\\\mylocalfolder\";
scnr = new Scanner(System.in);
}
public void askForWhichDetails(String fileName) throws IOException {
FileInputStream fileByteStream = null; // File input stream
Scanner inFS = null; // Scanner object
String textLine = null;
ArrayList aList1 = new ArrayList();
int i = 0;
int option = 0;
boolean bailOut = false;
// Try to open file
fileByteStream = new FileInputStream(filePath + fileName + \".txt\");
inFS = new Scanner(fileByteStream);
while (inFS.hasNextLine() && bailOut == false) {
textLine = inFS.nextLine();
if (textLine.cont.
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Let (X,T) be a topological space,.A subset X. Suppose that for all x epsilon A, U epsilon T s.t. x
epsilon U subset A. Show that.4 is open in X.
Solution
take the union of all such open sets
i.e. Union of U_x for all x in A
Union U_x is open since U_x is open for all x.
and Union U_x = A (Union U_x is a subset of A and A is subset of Union U_x)
So A is open.
Choose the best possible answer for each of the following multiple ch.pdf
Choose the best possible answer for each of the following multiple choice problems. Each
answer is worth Anthropology is a diverse discipline. Choose the combination of subfields that
best describes anthropology in the United States: Socio/cultural anthropology, biological
anthropology, linguistic anthropology. Socio/cultural anthropology, biological anthropology.
Socio/cultural anthropology, biological anthropology, linguistic anthropology, archaeology.
Socio/cultural anthropology, biological anthropology, archaeology. What are physical
anthropologists studying: The behavior, ecology, and culture of nonhuman primates. Culture in
humans and nonhuman primates. The biology of humans within the framework of evolution with
an emphasis on the interaction between biology and culture. Human speech within the
framework of evolution. Physical anthropologists follow the scientific method to produce new
knowledge. Which sequence of scientific inquiry is correct: collect data in the lab or in the wild
identify a problem/question or state a hypothesis related to a phenomenon observe a
phenomenon in the natural world test a hypothesis analyze the data discuss the research findings
in a publication (1) (3) (5) (2) (4) (6) (3) (2) (1) (5) (4) (6) (2) (3) (1) (4) (6) (5) (3) (1) (4) (5)
(2) (6) Charles R. Darwin is usually credited with the discovery of how new species can arise
from existing ones, but who was the other scientist who likewise discovered this principle of
evolution at around the same time as Darwin? August Weismann Alfred Russel Wallace Jean-
Baptiste de Lamarck John Ray
Solution
21. c
Explanation: Anthropology is a very diverse discipline. It is related to studies on various aspects
of human beings. These aspects can be social, cultural, related to biology, related to language etc.
So, option \'c\' stating four sub-disciplines is correct.
22. c
Explanation: Physical anthropology is a branch of anthropology. It is also known as biological
anthropology. This discipline relates to biological and behavioral aspects of human beings. These
include human physiology, anatomy, behavioral ecology of humans, primatology, population
genetics etc.
23. b
Explanation: First you need to observe a phenomenon in the natural world. Then identify a
problem, test a hypothesis and in the end, the research findings are published.
24. b
Explanation: Wallace was the scientist who discovered the principle of evolution at the same
time along with Darwin. Please note that Lamarck was the first one to discover this principle,
and had done the work before Darwin and Wallace..
If T is a bounded and self-adjoint operator on a Hilbert space and T_.pdf
If T is a bounded and self-adjoint operator on a Hilbert space and T_2 = T, show that T is the
orthogonal projection onto its range.
Solution
Since T is bounded and also self joint we have
T = T* and T* = T
Since TT* = T*T
we get TT = T*T*
i.e. T is normal.
So it follows that T is the orthogonal projection onto its range..
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