1. Halobacterium salinarum requires high salt concentrations to surv.pdf

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1. Halobacterium salinarum requires high salt concentrations to survive. When kept in low salt solutions, the cells swell due to influx of water from the surrounding and the cells eventually burst. Solution 1. Halobacterium salinarum requires high salt concentrations to survive. When kept in low salt solutions, the cells swell due to influx of water from the surrounding and the cells eventually burst..

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Solution Net cash provided by operating activities for the year ended December 31, 2014 Eliminate Depreciation expense +80000 Adjust revenue and expenses from accrual accounting to cashNet income151490 Eliminate Depreciation expense +80000 Adjust revenue and expenses from accrual accounting to cashIncrease in account receivable(20370)Decrease in inventory+12840Increase in prepaid expenses(4370)Decrease in account payable(6480)Increase in accrued expenses payable+11240Total cash generated by operating activities224360.

SolutionNet cash provided by operating activities for the year end.pdf

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SOA ( Service Oriented Architecture): SOA is a type of architectural pattern in computer software design where application components provides services to other components via network.This removes the need for third party product,vendor or technology. The main functionality to this type of model is to make it easier for software components connected over the internet to communicate with each other without human interaction.so every computer can run any number of services and this can be archeived without changing the main program itself.SOA provides a cost effective way to evolve and enhance Enterprise information systems. Web 2.0 on the other hand is world wide web websites that emphasize user-generated content,interoperability for end users. Some have agreed that web 2.0 is the best form of realizing SOA. 1)-2) Interoperability and portability: SOA can be realized through the help of webservices, because the best characterstic of webservices is the interoperability.Different distributed web services run on variety of software platforms and hardware architectures.JAVA j2EE and java WSDP are the best examples. 3) Platform independence: For example take a case where XML,XML schema are there which are used to publish,describe,search information. and there are multiple technologies and versions like SOAP, WSDL, UDDI, and ebXML to do this. For the web services to be functional, we need an environment of platform independent model.For example consider AJAX (asynchronous javascript and XML) is not a technology, it is just a technique to createinteractive webpages for this it supports XHTML,CSS,DOM,XML,XSLT,ATOM,SQL,REST,SOAP etc and uses javascript to glue it all together.This is a perfect example of SOA present in current trend. Solution SOA ( Service Oriented Architecture): SOA is a type of architectural pattern in computer software design where application components provides services to other components via network.This removes the need for third party product,vendor or technology. The main functionality to this type of model is to make it easier for software components connected over the internet to communicate with each other without human interaction.so every computer can run any number of services and this can be archeived without changing the main program itself.SOA provides a cost effective way to evolve and enhance Enterprise information systems. Web 2.0 on the other hand is world wide web websites that emphasize user-generated content,interoperability for end users. Some have agreed that web 2.0 is the best form of realizing SOA. 1)-2) Interoperability and portability: SOA can be realized through the help of webservices, because the best characterstic of webservices is the interoperability.Different distributed web services run on variety of software platforms and hardware architectures.JAVA j2EE and java WSDP are the best examples. 3) Platform independence: For example take a case where XML,XML schema are there which are used to publis.

SOA ( Service Oriented Architecture)SOA is a type of architectura.pdf

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C. 2nd. .pdf

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pH= -log(1x10-8)= 8SolutionpH= -log(1x10-8)= 8.pdf

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P0 = 21.40 D1 = 1.07 D2 =1.1449 D3 = 1.2250 P3 = 26.22 Growth rate in dividends: PV= 1.07; FV = 1.2240; N = 2; CPT I/Y; I/Y = 6.9544% Let\'s round to 7%. Dividend yield = (d1/P0) = 1.07/21.40 = 0.05 Total return = 7% + 5% = 12% Solution P0 = 21.40 D1 = 1.07 D2 =1.1449 D3 = 1.2250 P3 = 26.22 Growth rate in dividends: PV= 1.07; FV = 1.2240; N = 2; CPT I/Y; I/Y = 6.9544% Let\'s round to 7%. Dividend yield = (d1/P0) = 1.07/21.40 = 0.05 Total return = 7% + 5% = 12%.

P0 = 21.40D1 = 1.07D2 =1.1449D3 = 1.2250P3 = 26.22Growth r.pdf

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Nt = No er t where density at time t No= = starting density Nr=dN/dt so r=1/N*dN/dt t =1 1):- Nt = No er t if No =1 =1* e-.1*1 =9.047 2):- Nt = No er t if No=3 =3* e0.1*1=1.105*3=3.3156 3):-Nt = No er t if No=7= 7*e0.37*1=10.08 4):-Nt = No er t if No=14 =14*e-0.3*1=10.36 Solution Nt = No er t where density at time t No= = starting density Nr=dN/dt so r=1/N*dN/dt t =1 1):- Nt = No er t if No =1 =1* e-.1*1 =9.047 2):- Nt = No er t if No=3 =3* e0.1*1=1.105*3=3.3156 3):-Nt = No er t if No=7= 7*e0.37*1=10.08 4):-Nt = No er t if No=14 =14*e-0.3*1=10.36.

Nt = No er twhere density at time tNo= = starting densityNr=dN.pdf

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Normalization is necessay step in creation of database design becaus.pdf

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Hi, Please find my codeimport java.util.Random;public class Pro.pdf

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For sound intensity, ratio in dB is given by: x = 10*log(I/Io) We know, I = P/(4*pi*D^2) <----- I is inversely proportional to D So, x = 10*log(D^2/(D+d)^2) where D = original distance from speaker d = distance he has moved from original distance from speaker So, -14 = 10*log((D/(D+d))^2) So, (D+d)/D = 5.01 So, 1 + d/D = 5.01 So, d/D = 4.01 So, d = 4.01*D <-------- So, he is about 4.01 times D away from original distance or 5.01*D distance away from speaker So, answer is : 5.01*D Solution For sound intensity, ratio in dB is given by: x = 10*log(I/Io) We know, I = P/(4*pi*D^2) <----- I is inversely proportional to D So, x = 10*log(D^2/(D+d)^2) where D = original distance from speaker d = distance he has moved from original distance from speaker So, -14 = 10*log((D/(D+d))^2) So, (D+d)/D = 5.01 So, 1 + d/D = 5.01 So, d/D = 4.01 So, d = 4.01*D <-------- So, he is about 4.01 times D away from original distance or 5.01*D distance away from speaker So, answer is : 5.01*D.

For sound intensity,ratio in dB is given byx = 10log(IIo)We.pdf

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Exp:- Listeria monocytogenes is a facultative anaerobic bacteria that can survive both aerobic and anaerobic conditions and is responsible for listeriosis disease. The bacterium can undergo fermentation in absence of oxygen and normal glycolysis and Krebs cycle.The main route of infection is ingestion of contaminated food. Solution Exp:- Listeria monocytogenes is a facultative anaerobic bacteria that can survive both aerobic and anaerobic conditions and is responsible for listeriosis disease. The bacterium can undergo fermentation in absence of oxygen and normal glycolysis and Krebs cycle.The main route of infection is ingestion of contaminated food..

Exp- Listeria monocytogenes is a facultative anaerobic bacteria tha.pdf

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DNA controversy is a dispute about whether Rosalind Franklin and Wilkins were given proper credit for their contribution to the determination of the structure of DNA. In January 1951, Franklin discovered that there were two forms of DNA by using x-ray diffraction techniques, she worked as a research associate at King\'s College London in the Medical Research Council\'s Biophysics Unit. The controversy arose from claims that some of the data were shown to Watson and Crick, without her knowledge. Her experimental results provided estimates of the water content of DNA crystals and these reults were consistent with the two sugar-phosphate backbones being on the outside of the molecule. Franklin personallytold Crick and Watson that hte X-ray diffraction images collected where the best evidence for the helical nature od DNA. Crick and Waatson had been to the public seminar presented by Franklin, where they had spoken to Wilkins about offering him a co-authorship on the article that first described the double helix structure of DNA. On the completion of their model, Crick and Watson had invited Wilkins to be a co-author of the paper describing the structure, he turned down this offer as he had taken no part in building the model. Watson, Francis Crick and Maurice Wilkins were awarded the 1962 Nobel Prize for Physiology or Medicine for their discoveries concerning the molecular structure of nucleic acids and its significance for information transfer in living material, with the help of the research of Rosalind Franklin. There is a little doubt that the significance of Franklin\'s data should have been acknowledged- the evidence is overwhelming that it played an important role in Watson and Crick\'s modelling process. Franklin could not share in the 1962 Nobel prize awarded to Watson and Crick because she had died of ovarian cancer in 1958. Hence, it cannot be concluded that Watson and Crick were not ethical about their research and also that they donot deserve a nobel prize for their discovery. And the Historical records contain all the data of the work presented by Franklin and also by Watson and Crick. Solution DNA controversy is a dispute about whether Rosalind Franklin and Wilkins were given proper credit for their contribution to the determination of the structure of DNA. In January 1951, Franklin discovered that there were two forms of DNA by using x-ray diffraction techniques, she worked as a research associate at King\'s College London in the Medical Research Council\'s Biophysics Unit. The controversy arose from claims that some of the data were shown to Watson and Crick, without her knowledge. Her experimental results provided estimates of the water content of DNA crystals and these reults were consistent with the two sugar-phosphate backbones being on the outside of the molecule. Franklin personallytold Crick and Watson that hte X-ray diffraction images collected where the best evidence for the helical nature od DNA. Crick and Waatson had .

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consider the cross between White petals and dark blue petals Flower color in sweet peas. Sweet peas - is the classic object of genetic research of traits inheritance. When scientists studied the color of flowers of this plant, they discover, that in some crosses the ratio of phenotype was 9:7. That is, from the crossing of two pure lines of sweet peas with white flowers, all plants in first-generation had the purple flowers. And from crossing of these hybrids was obtained the offspring with the phenotypic ratio : 9 purple: 7 white. If the ratio was 3:1, we could suppose, that we are dealing with a simple monohybrid crossing, but in this case the ratio is differ. So obviously, the genetic trait of color of sweet peas flowers is controlled by two nonallelic genes, which interact each other. For convenience, we mark these genes as \"C\" and \"P\". Thus, the pure lines will have genotypes \"CCpp\" and \"ccPP\", and first generation hybrids respectively genotype \"CcPp\". The mechanism of genetic interaction of these genes has become known recently. It was found that the color of sweet pea flowers depends on the synthesis of pigments - anthocyanins. The synthesis of anthocyanins occurs in two stages. Gene \"C\" is responsible for the first stage of the synthesis, while the gene \"P\" for second, and even if only one dominant allele from these genes is lack, then synthesis of anthocyanins is not occurs. That is to say, that the presence of a recessive allele in the homozygous state in the genotype of either of the two genes is blocks the development of trait of flowers coloration. Such interaction of nonallelic genes is called a double recessive epistasis. We can describe this interaction like this - recessive allele of each gene in the homozygous state suppresses the expression of a dominant allele of another ( \"cc\" suppresses \"P\", \"pp\" suppresses \"C\" ). Genetic traits file for this case should be like this : Solution consider the cross between White petals and dark blue petals Flower color in sweet peas. Sweet peas - is the classic object of genetic research of traits inheritance. When scientists studied the color of flowers of this plant, they discover, that in some crosses the ratio of phenotype was 9:7. That is, from the crossing of two pure lines of sweet peas with white flowers, all plants in first-generation had the purple flowers. And from crossing of these hybrids was obtained the offspring with the phenotypic ratio : 9 purple: 7 white. If the ratio was 3:1, we could suppose, that we are dealing with a simple monohybrid crossing, but in this case the ratio is differ. So obviously, the genetic trait of color of sweet peas flowers is controlled by two nonallelic genes, which interact each other. For convenience, we mark these genes as \"C\" and \"P\". Thus, the pure lines will have genotypes \"CCpp\" and \"ccPP\", and first generation hybrids respectively genotype \"CcPp\". The mechanism of genetic interaction of these genes has bec.

consider the cross between White petals and dark blue petalsFlower.pdf

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Answer: .Ltext0: .section .rodata .LC0: 0000 0A0A0949 .string \"\ \ \\tIN DECIMAL FORM: %d\" .text .globl convert1 convert1: .LFB0: .cfi_startproc 0000 55 pushq %rbp .cfi_def_cfa_offset 16 .cfi_offset 6, -16 0001 4889E5 movq %rsp, %rbp .cfi_def_cfa_register 6 0004 4883EC30 subq $48, %rsp 0008 48897DD8 movq %rdi, -40(%rbp) 000c 488975D0 movq %rsi, -48(%rbp) 0010 C745F800 movl $0, -8(%rbp) 0017 C745FC01 movl $1, -4(%rbp) 001e C745F020 movl $32, -16(%rbp) 0025 C745EC0F movl $15, -20(%rbp) 002c EB48 jmp .L2 .L6: 002e 836DF001 subl $1, -16(%rbp) 0032 8B45EC movl -20(%rbp), %eax 0035 4898 cltq 0037 488D1485 leaq 0(,%rax,4), %rdx 003f 488B45D8 movq -40(%rbp), %rax 0043 4801D0 addq %rdx, %rax 0046 8B00 movl (%rax), %eax 0048 83F801 cmpl $1, %eax 004b 7525 jne .L3 004d C745FC01 movl $1, -4(%rbp) 0054 C745F401 movl $1, -12(%rbp) 005b EB07 jmp .L4 .L5: 0060 8345F401 addl $1, -12(%rbp) 005d D165FC sall -4(%rbp) .L4: 0064 8B45F4 movl -12(%rbp), %eax 0067 3B45F0 cmpl -16(%rbp), %eax 006a 7EF1 jle .L5 006c 8B45FC movl -4(%rbp), %eax 006f 0145F8 addl %eax, -8(%rbp) .L3: 0072 836DEC01 subl $1, -20(%rbp) .L2: 0076 837DEC00 cmpl $0, -20(%rbp) 007a 79B2 jns .L6 007c C745FC01 movl $1, -4(%rbp) 0083 C745EC0F movl $15, -20(%rbp) 008a EB44 jmp .L7 .L11: 008c 8B45EC movl -20(%rbp), %eax 008f 4898 cltq 0091 488D1485 leaq 0(,%rax,4), %rdx 0099 488B45D0 movq -48(%rbp), %rax 009d 4801D0 addq %rdx, %rax 00a0 8B00 movl (%rax), %eax 00a2 83F801 cmpl $1, %eax 00a5 7525 jne .L8 00a7 C745FC01 movl $1, -4(%rbp) 00ae C745F401 movl $1, -12(%rbp) 00b5 EB07 jmp .L9 .L10: 00ba 8345F401 addl $1, -12(%rbp) 00b7 D165FC sall -4(%rbp) .L9: 00be 8B45F4 movl -12(%rbp), %eax 00c1 3B45EC cmpl -20(%rbp), %eax 00c4 7EF1 jle .L10 00c6 8B45FC movl -4(%rbp), %eax 00c9 0145F8 addl %eax, -8(%rbp) .L8: 00cc 836DEC01 subl $1, -20(%rbp) .L7: 00d0 837DEC00 cmpl $0, -20(%rbp) 00d4 79B6 jns .L11 00d6 8B45F8 movl -8(%rbp), %eax 00d9 89C6 movl %eax, %esi 00db BF000000 movl $.LC0, %edi 00e0 B8000000 movl $0, %eax 00e5 E8000000 call printf 00ea 90 nop 00eb C9 leave .cfi_def_cfa 7, 8 00ec C3 ret .cfi_endproc .LFE0: .globl convert2 convert2: .LFB1: .cfi_startproc 00ed 55 pushq %rbp .cfi_def_cfa_offset 16 .cfi_offset 6, -16 00ee 4889E5 movq %rsp, %rbp .cfi_def_cfa_register 6 00f1 897DEC movl %edi, -20(%rbp) 00f4 488975E0 movq %rsi, -32(%rbp) 00f8 C745F800 movl $0, -8(%rbp) 000000 00ff 8B45EC movl -20(%rbp), %eax 0102 8945FC movl %eax, -4(%rbp) 0105 EB3F jmp .L13 .L15: 0107 8B45FC movl -4(%rbp), %eax 010a 99 cltd 010b C1EA1F shrl $31, %edx 010e 01D0 addl %edx, %eax 0110 83E001 andl $1, %eax 0113 29D0 subl %edx, %eax 0115 8945EC movl %eax, -20(%rbp) 0118 8B45F8 movl -8(%rbp), %eax 011b 8D5001 leal 1(%rax), %edx 011e 8955F8 movl %edx, -8(%rbp) 0121 4898 cltq 0123 488D1485 leaq 0(,%rax,4), %rdx 012b 488B45E0 movq -32(%rbp), %rax 012f 4801C2 addq %rax, %rdx 0132 8B45EC movl -20(%rbp), %eax 0135 8902 movl %eax, (%rdx) 0137 8B45FC movl -4(%rbp), %eax 013a 89C2 movl %eax, %edx 013c C1EA1F shrl $31, %edx 013f 01D0 addl .

Answer .Ltext0 .section .rodata .LC0 0000.pdf

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