What type of epistasis between locus A and B could be explained by t.pdf

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What type of epistasis between locus A and B could be explained by the following biochemical pathway? Select one: single recessive epistasis dominant/recessive epistasis single dominant epistasis duplicate dominant epistasis duplicate genes with cumulative effect duplicate recessive epistasis Solution What type of epistasis between locus A and B could be explained by the following biochemical pathway dominant/recessive epistasis---answer =================================== epistasi : it can referred as the interaction of genes that are not alleles, in specific the suppression of the effect of one such gene by another.

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For a given class, the studen records are stored in a file. The records are randomly accessed and updated. Assume that each student\'s record is of fixed size. Which of the three allocation schemes (contiguous, linked and table/indexed) will be the most appropriate? Solution As each file has its own index block which stores the addresses of memory occupied by the file so Table/indexed allocation scheme would be most appropriate as we can easily access any block with help of index,.

For a given class, the studen records are stored in a file. The reco.pdf

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\"Elementary school children spent significantly more time reading in the presence of a teacher than when the teacher was absent, t(7) = 2.804, p < 0.05 (2 = 0.46).\" In this related samples t test, how many participants were observed? Solution t(7) means df = 7, and participants must be 7+1 = 8 (unless the control group students were identical twins of those in experiment, making 8*2=16 participants in dependent samples).

Elementary school children spent significantly more time reading i.pdf

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Describe the tools and technology used to support IT project management. What are the characteristics of future tools? Solution Project management is defined as the application of a collection of tools and techniques which are used to direct the use of the resources toward the accomplishment of a unique, complex, one- time task.This task has boundries that needs to be taken care of such as the project must be completed within time limit, cost, and quality constraints. Each task requires a particular mix of these tools and techniques structured to fit the task environment and life cycle of the task. The project management techniques related to the project initiation phase include: [16].

Describe the tools and technology used to support IT project managem.pdf

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Creat Shape classes from scratch DETAILS You will create 3 shape classes (Circle, Rectangle, Triangle) that all inherit from a single abstract class called AbstractShape which implements Shape (also created by you). You are also responsible for creating the driver class \"Assignment7.java\" (program that tests your classes and described on page 3) which does the following reads input data from a file instantiates various objects of the three shapes based on the input data stores each in a LinkedList outputs this list to an output file sorts a \"copy\" of this LinkedList of objects outputs the sorted version of the list to the output file outputs the original list to the output file This driver program also needs to \"ignore errors in the input file that breach the specified input format as described in the Assianment7,java details (see page 3 1. Shape.java This is an interface that has 2 abstract methods, and passes the responsibility of implementing the compareTo method to the class that implements Shape (you may note, nomally Comparable is \"implemented\" by a class. However, an interface cannot implement because interfaces can only contain abstract methods. That said, an interface can only extend other interfaces and the responsibility of actually \"implementing\" the abstract method(s) of the super class interface is passed on to the sub-classes) public interface Shape extends Comparable public double calculateAreal) Il This abstract method is implemented at the concrete level public Shape copyShape); Il also implemented at the concrete level 2. AbstractShape.java public abstract class AbstractShape implements Shape This class should contain an instance field to store the name of each obiect. The constructor which sets this field should receive the name and a number to be concatenated to the name and then stored in the name field Recall, when the super class has a parameterized constructor, the sub-classes will need to call it AND the sub- classes will need to also provide a constructor without parameters This abstract class will implement the compareTo method passed on from the Shape interface and will pass on the responsibility of implementing calculateArea to the extending sub-classes (compare To will use the calculateArea method when comparing 2 Shape objects). Along with compare To, one more concrete method should be included. The following will be used by the sub-classes\' toString method: public String getName) II Simply returns the name field data Solution in7.txt 4.4 2.5 3 8.1 3.0 5.0 2.5 3 4 2.5 tuesday -7 1.0 3 three 3 -9 3 5 1.0 Assignment7.java import java.io.*; import java.util.*; public class Assignment7 { /** * This is the test driver class that will include main. * This program MUST read a file named in7.txt and * generate an output file named out7.txt. The in7.txt * file must be created by you based on formatting * described shortly. * * @param theArgs */ public static void main(String[] theArgs) { List myList = new Arra.

Creat Shape classes from scratch DETAILS You will create 3 shape cla.pdf

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Compare the following: bacterial species, bacterial strain, and bacterial type. Explain the relationship and development of strains and types to a species. Solution Bacterial species: It is a group of prokaryotic organisms which are ecologically distinct where many ecotypes of bacteria are grouped under a genous. These bacteria show a certain level of phenotypic identity, above 95% pair-wise average nucleotide identity (ANI) and over 97% 16S rRNA identity. bacterial strain: A strain is a low level taxonomically characterized micro-organism within a particular species that show genetic variance. Bacterial types: Bacteria can be classified based on cell wall type, and shape into different types. Based on cell wall type: gram positive, cells take up purple gram stain due to peptidoglycan rich cell wall egs (Streptococcus pneumoneae). Gram-negative bacteria such as vibrio cholerae which cause cholera take up pink stain due to thick peptidoglycan layer in their cell wall. Based on shape bacteria are classified into coccus (spherical shaped bacteria, egs.streptococcus), bacillus (rod-shaped bacteria egs. Ecoli), Vibrio (spiral or comma-shaped bacteria egs. Vibrio cholerae), spirillum (Spiral shape bacteria with more rigid structure egs: Heliobacter Pylori corkscrew shaped bacteria), Spirochete (Long thin flexible bacteria which are corkscrew shaped egs: Treponema pallidum) A species is a higher taxonomical classification which further breaks down into different strains that are genetically different. A type of bacteria shows which morphological differences and cell wall characteristics of a bacteria. The lowest level of classification would be based on type of species. A group of organisms of the different type, different morphological feature and virulence carrying out similar function would be classified under a strain. These strain would have micro-organisms, with genetic variance egs. A species of bacteria E. coli lives in the intestine and is good for human but it can also turn out to be pathogenic due to the 5% genetic variance among strains. Enteropathogenic and enterotoxigenic bacteria cause diarrhea in humans particularly children. Many such strains are grouped together under species egs E. Coli species. Which contains all the pathogenic and nonpathogenic E.Coli bacteria. These bacteria have 97% 16s rRNA identity, therefore falling under one species..

Compare the following bacterial species, bacterial strain, and bact.pdf

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Annual estimates of the population in a Hawaiian county from 1999 (t = 1) onward are shown in the time series graph below. What is the dominant component in this time series? Click the icon to view the time series plot. Solution The dominant part in this time series is just the upward trend. It is not cyclic (repetative), it is not seasonal (some time up and some time down) and it is not random either (not that random actually)..

Annual estimates of the population in a Hawaiian county from 1999 (t .pdf

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Acid rain negatively affects plants by A. reducing the photosynthetic rate in the leaves B. burning the leaves and bark. C. leaching cations that the plant needs from the soil D. leaching anions from the soil that the plant needs Solution Acid rain is a mixture of inorganic acids like H2SO4 and HNO3. These acids are from upon the reaction of pollutant like NOx and SO2 with water oxygen and other components present in the air. Both the acids are strong acid and it is the property of strong acid to dissociate into proton and Sulphate or Nitrate ions. In the soil cation, such as Potassium, calcium and magnesium are present, in their mineral forms, which are exchange with proton in acidic condition. Once cation is exchanged they leach into the soil are no more available for the plants..

Acid rain negatively affects plants byA. reducing the photosynthet.pdf

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A random variable X is exponentially distributed with a mean of 0.14.pdf

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A protein, called pRb (protein for retinoblastoma) is synthesized by a gene (Rb) on chromosome 13 in humans. When mutant or absent in the homozygous state in a retinal cell, there is a high likelihood of developing a cancerous condition of the eye called retinoblastoma. Nonmutant pRb binds with and temporarily inactivates E2F, a transcription factor that stimulates genes to produce proteins that initiate S phase of the mitotic cell cycle (thus, Rb is a negative regulator of cell division). When phosphorylated in response to a variety of normal stimuli, pRb normally releases E2F and cell division is stimulated. Thus, Rb is a tumor suppressor gene. (For a recent review, see H. Liu et al. (2004), \"Novel functions of the RB tumor suppressor.\" Current Opinions in Genetics and Development 14: 55–64. When retinoblastoma occurs in families, the disease tends to have an early onset and express bilaterally (both eyes). When sporadic (spontaneous, not inherited), it tends to show later onset and be unilateral (one eye only). Individuals who inherit a predisposition for the disease are Rb+/Rb– and may get retinoblastoma by having a mutation in the Rb+ allele.Assume that two heterozygous parents have children. One would expect the Rb–/Rb– genotypic combination in a zygote (one-cell stage of an individual) to be characterized as: a.lethal (inviable or lasting a relatively short time) b.completely normal c.having a very low likelihood of developing retinoblastoma d.having unilateral retinoblastoma only Solution A. the Rb-/Rb- genotype B. the Rb+/Rb- genotype C. the Rb+/Rb+ genotype So answer is A.

A protein, called pRb (protein for retinoblastoma) is synthesized by.pdf

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What kind of bond would form between NH_4 and Cl^- to make the salt ammonium chloride (NH_4Cl)? Iconic. Polar covalent. Hydrogen. Nonpolar covalent. A or C, depending on conditions. When two atoms are equally electronegative, they will interact to form a hydrogen bond. nonpolar covalent bond. pair of ions. ionic bond. polar covalent bond. A(n) occurs between nonpolar molecules. ionic attraction van der force hydrogen bond nonpolar covalent bond Which of the following statements about the density of water is accurate? dense than water vapor. Liquid water is denser than ice, which is more water behaves like other substances in terms of density. At all temperatures, water vapor. liquid water, which is water vapor is denser than ice, which is more dense than liquid water. Solution Answer: Answer: 6). a). Ionic NH4Cl, or Ammonium Chloride, is a salt, which means it has an ionic bond. 7). d). nonpolar covalent bond Nonpolar covalent bonds are a type of chemicalbond where two atoms share a pair of electrons with each other. Polar covalent bonding is a type of chemical bond where a pair of electrons is unequally shared between two atoms. 8). c). van der Waals force If the molecule is nonpolar, then no dipole-dipole interactions or hydrogen bonding can occur and the only possible intermolecular force is the weak van der Waals force 9). a). Liquid water is more dense than ice, which is more dense than water vapor When water freezes, water molecules form a crystalline structure maintained by hydrogen bonding. Solid water, or ice, is less dense than liquid water.Ice is less dense than water because the orientation of hydrogen bonds causes molecules to push farther apart, which lowers the density. In the gas/vapor phase, molecules have very little interaction with each other and they are at large distance, whereas in liquid and solid, molecules are in continuous contact with each other. It is therefore clear that there are more molecules per liter in ice than in water vapor, and therefore the density is larger..

What kind of bond would form between NH_4 and Cl^- to make the salt a.pdf

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Which muscle provides a guide to the position of the radial artery at the wrist fr taking the pulse? Solution answer is D this Flexor Carpi Radialis which runs diagonally across forearm; midway, its fleshy belly replaced by a flat tendon that becomes cordlike at wrist, O-medial epicondyle of humerus I- base of second and 3rd metacarpals, insertion tendon easily seen and provides guide to position of radial artery (Used for pulse taking) at wrist.

Which muscle provides a guide to the position of the radial artery at.pdf

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Which of the following is a true statement regarding lysosomes? Which of the following is a true statement regarding lysosomes?A. The lysosome may mature from a late endosome.B. The contents of the lysosome originate in the smooth ER.C. The lysosome may function in glycosylation.D. The lysosome functions in export packaging.E. None of these are correct. Solution A) The lysosome may mature from a late endosome. Early endosomes mature into late endosomes, which in turn mature into lysosomes..

Which of the following is a true statement regarding lysosomesWhi.pdf

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When a signal arrives at the cell, it gets transduced. What is meant.pdf

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what is application of any position sensor how it worksSolut.pdf

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What are two principal categories of SkimmingSolutionAns) Skim.pdf

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There are many ways to realizing a second-order system. The following circuit, incorporating an operational amplifier configured as an inverting integrator, is an attractive implementation. This circuit is not limited by the amplifier\'s common-mode input voltage range and its performance is good at high frequencies. Determine the transfer fucntion, that is, the ratio of output to input, for this circuit. Solution Op amplifier can be used as a inverting or non inverting mode. Where inverting integrator is designed by applying input to the inverting terminal and output is taken across capacitor.

There are many ways to realizing a second-order system. The followin.pdf

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The nuclear genome of a mouse nerve cell is compared to that of a mo.pdf

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The pedigree below represents a family with a fairly common hereditar.pdf

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The diffusion length for electrons L_n = 3.16 mu in an n-doped (N_d .pdf

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1) (a) Write a code fragment which adds the 32-bit contents of $8100 (LSB) - 8103 (MSB) to $8104 - $8107 and stores it to $8100 - $8103. (b) Write a similar code fragment, which adds the 8-digit BCD contents of $8200 - $8203 to $8204 - $8207 and stores it to $8200 - $8203. Using 68HC11 Assembly Instruction Set. Solution Step I : Initialize the data segment. Step II : Load the LSB of first number into AX register. Step III : Load the MSB of first number into BX register. Step IV : Load the LSB of the second number into CX register. Step V : Load the MSB of the second number into DX register. Step VI : Add the LSBs of two number. Step VII : Add the MSBs of two numbers along with carry. Step VIII : Display the result. Step IX : Stop. .model small .data op1 dd 12345678h op2 dd 11111111h Corel - 10 ans dd ? .code mov ax, @data mov ds, ax mov ax, word ptr op1 ; lsb of number1 in ax mov bx, word ptr op1+2 ; msb of number1 in bx mov cx, word ptr op2 ; lsb of number2 in cx mov dx, word ptr op2+2 ; msb of number1 in dx add ax, cx ; add msb + msb + carry mov word ptr ans, ax ; lsb answer mov word ptr ans+2, bx ; msb answer mov bx, word ptr ans+2 ; Result in reg bx mov dh, 2 l1: mov ch, 04h ; Count of digits to be displayed mov cl, 04h ; Count to roll by 4 bits l2: rol bx, cl ; roll bl so that msb comes to lsb mov dl, bl ; load dl with data to be displayed and dl, 0fH ; get only lsb cmp dl, 09 ; check if digit is 0-9 or letter A-F jbe l4 add dl, 07 ; if letter add 37H else only add 30H l4: add dl, 30H mov ah, 02 ; INT 21H (Display character) int 21H dec ch ; Decrement Count jnz l2 dec dh cmp dh, 0 mov bx, word ptr ans ; display lsb of answer jnz l1 mov ah, 4ch ; Terminate Program int 21h end 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 .model small .data op1 dd 12345678h op2 dd 11111111h Corel - 10 ans dd ? .code mov ax, @data mov ds, ax mov ax, word ptr op1 ; lsb of number1 in ax mov bx, word ptr op1+2 ; msb of number1 in bx mov cx, word ptr op2 ; lsb of number2 in cx mov dx, word ptr op2+2 ; msb of number1 in dx add ax, cx ; add msb + msb + carry mov word ptr ans, ax ; lsb answer mov word ptr ans+2, bx ; msb answer mov bx, word ptr ans+2 ; Result in reg bx mov dh, 2 l1: mov ch, 04h ; Count of digits to be displayed mov cl, 04h ; Count to roll by 4 bits l2: rol bx, cl ; roll bl so that msb comes to lsb mov dl, bl ; load dl with data to be displayed and dl, 0fH ; get only lsb cmp dl, 09 ; check if digit is 0-9 or letter A-F jbe l4 add dl, 07 ; if letter add 37H else only add 30H l4: add dl, 30H mov ah, 02 ; INT 21H (Display character) int 21H dec ch ; Decrement Count jnz l2 dec dh cmp dh, 0 mov bx, word ptr ans ; display lsb of answer jnz l1 mov ah, 4ch ; Terminate Program int 21h --------------------------------------------------- Step 1 : Initialize the data segment. Step 2 : Get the first number in AL register. Step 3 : Get the second number in BL register. Step 4 : Add the two numbers. Step .

1) (a) Write a code fragment which adds the 32-bit contents of $8100.pdf

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