Which of the following protons would you NOT expect to find embedded .pdf

•

0 likes•2 views

Which of the following protons would you NOT expect to find embedded in the cytoplasmic membrane of any bacteria? Primary active transport proteins. ATP synthase. Enzyme I complex of the Group translocation system. Basel body protein rings of flagella. Which of the following is true? Both photosystem I and photosystem II rely on light bombarding chlorophyll. Photosystem II donates electrons to photosystem I in a linear electron transport chain. Photosystem I only (not photosystem II) directly produces NADPH utilized for the Calvin cycle. All of the above are true. None of the above are true. Which one of these prokaryotic characteristics can\'t co-exist in a single cell? Mesospheric and acidophil. Aero-tolerant and basophilic. Thermophiles and psychrophilic. Fermentative and Anaerobic. If a population of 4 bacterial cells in log phase, how many cells can we expect to see in an hour\'s time if the cell has a generation time of 30 minutes? 8 cells. 16 cells. 64 cells. 128 cells. The classification of ribosomes in bacteria would be: Ribosomes are proteins and bacteria contain 70s versions. Ribosomes are proteins and bacteria contain 80s versions. Ribosomes are membrane lipids and bacteria contain 70s versions. Ribosomes are membrane lipids and bacteria contain 80s versions. Where do The Calvin Cycle and Glycolysis converge? Only after the \"prep step\" once acetyl CoA is formed, leading to the Calvin Cycle Only when NADH is utilized in the electron transport chain At Glyceraldehyde 3 Phosphate. None of the above are correct. Solution 8.a primary active transport proteins The basal body of a bacterial flagellum is a rod and a system of rings embedded in the cell envelope.Gram-negative flagella usually have an L ring in the plane of the lipopolysaccharide in the outer membrane,a periplasmic P ring in the plane of the peptidoglycan around the flagellar rod and a M S ring that is located within and above the cytoplasmic membrane.A ,C ring extends into the cytoplasm. Enzyme I transfers phosphate from PEP to HPr.PEP group translocation is a distinct method used by bacteria for sugar uptake where the source of energy is from phosphoenolpyruvate (PEP).It is known as multicomponent system that always involves enzymes of the plasma membrane and those in the cytoplasm. ATP synthase is located within the thylakoid membrane and the inner mitochondrial membrane in bacteria. 9.d all are true. In photosystem I energy is absorbed by a pair of P700 chlrophyll a molecules raise to an excited energy level.From there they pass onto FeS4,then onto ferrodoxin,and finally onto ferrodoxin- NADP reductase.After 2 electrons have reduced ferrodoxin-NADP reductase they are transferred to NADP+ reducing it to NADPH and a H+.Photosystem II works with Photosystem I and two series of enzymes imbedded in the thylakoid membrane to transfer energy from the form of light to that stored in chemical bands and gradients which the plant can use in a process called noncyclic photophosph.

Education

to NADP+ reducing it to NADPH and a H+.Photosystem II works with Photosystem I and two
series of enzymes imbedded in the thylakoid membrane to transfer energy from the form of light
to that stored in chemical bands and gradients which the plant can use in a process called
noncyclic photophosphorylation.It not only transfers the electrons for PS I but also is responsible
for ATP production.
10.c thermophilic-heat loving (50-70 degrees) and psychrophilic -cold loving (-10 to 25 degrees)
11.b 16 cells .
in one hour bacteria divides 2 times.22=4
4*4=16
12.a
13.b.

Similar to Which of the following protons would you NOT expect to find embedded .pdf

1.cell_-_organelles_transport_mechanism.pdf

SajidSSaiyad1

Chromosome & nuclear structure.pdf

sumitraDas14

Intracellular trafficking and protein sorting

apeksha40

Cell organelles

IIDC

Cell organelles

IIDC

Leeuwenhoek described microorganism as? Animalcules. Prokaryotes. Eukaryotes. Protozoa. Which of the following was not mentioned as a central pathway for glucose metabolism? Entner- Doudoroff pathway. The Pentose Phosphate way. Glycolysis. Fermentation. Answers A and B. In a health Gram negative cell, which of these would you expect to be the most exterior to the cell? Lipid A. O-side chain. ATP synthase. Secondary active transport proteins. Axial filaments would be associated with: Flagella on a bacillus. Biofilm construction. Spirochete motility. Gram negative outer membranes. N-Acetyl Glucosamine associates with which of the following? N- acetyl numeric acid only. N-acetyl numeric acid and N-acetyl glucosamine N-acetyl glucosamine and technical acid. N-acetyl numeric acid and lipotechoic acid. Leeuwenhoek examined rain and pond water for microorganisms. He found some bacteria in one sample while the other did not contain any life. The independent variable in this procedure was: The presence or absence of bacteria. Where the water was obtained from. How long the water was allowed to be examined on his scope. The possibility that the bacteria in a sample was introduced by the microscope itself. Solution 14. a (Animalcules) Leeuwenhoek observed micro-organisms for the first time in microscope. They appeared to be little animals to him so he called them animalcules. These descriptions were made for the first time which made him the first microbiologist. 15. b. Pentose Phosphate pathway Glycolysis and Fermentation are central pathways for carbohydrate metabolism for almost all aerobes and Anaerobic organisms. ED pathway uses enzymes like 6-phospho gluconate dehdrogenase and convert glucose to pyruvate in organisms like Pseudomonas. Pentose Phosphate pathway is an alternate glucose oxidation pathway to produce NADPH rather than being a central metabolic pathway, 16. b. O-specific polysaccharide chain O-specific poly sccharide chain is the outer most part of outer membrane of Gram Negative cell wall. Gram negative cell wall has has two membranes outer and inner separated by a periplasmic space filled with Lipoproteins. 17. c. Spirochaete motility Spirochetes are spiral cells contain an axial filament wrapped around them. They are similar to flagella and are used for motility of the organism. According to Chegg\'s policies only first 4 questions are to be answered. If you have multiple questions please post them as groups 4 questions related to each other..

Leeuwenhoek described microorganism as Animalcules. Prokaryotes. Euk.pdf

rbjain2007

Bio.. mcq..

Anand Messy

Role of endoplasmic reticulum in protein systhesis and1

Monalisa Behera

What are rubisco and RuBP? And what do they do? Briefly list 4 similarities in how ATP is made in mitochondria and in chloroplasts. A. How does the second law of thermodynamics explain why diffusion occurs across a membrane? B. Describe how oxidative phosphorylation, substrate-level phosphorylation, and photo phosphorylation differ in the way ATP is made. Solution RuBisCO , is an enzyme Which is used in the Calvin cycle for catalyzing the first step of carbon fixation, a process by which the atoms of atmospheric carbon dioxide are made available to organisms in the form of energy-rich molecules such as sucrose. RuBisCO catalyzes either the carboxylation or oxygenation of ribulose-1,5-bisphosphate (also known as RuBP) with carbon dioxide or oxygen. Ribulose-1,5-bisphosphate (RuBP) is an important 5-carbon intermediate in the Calvin cycle taking place during photosynthesis. It is the substrate, which is used by the enzyme to fix CO2 to create a highly unstable 6 Carbon PO4 which decays into two molecules of glycerate 3- phosphate. In both i.e. Mitochindria as well as in Chloroplast ATP synthesis takes place via a proton gradient- Both have ATP synthatases Both have Electron Transport Chains Both have 70s ribosomes. The second law is a trend towards the randimization or increasing entropy. When the concentration of a substance on both sides of a membrane are equal, the distribution is more random than when they are unequal. Diffusion of a substance to a region, where it is initially less concentrated increases entropy, making it energetically favourable (SPONTANEOUS) process. -Substrate level phosphorylation occurs in the cytoplasm during Glycolysis and Mitochondria during Krebs cycle. -2ATP/GTP are produced by conversion of ADP or GDP OXIDATIVE PHOSPHORYLATION -Occurs in Mitochondria Occurs during respiration pigment systems are not involved ATP is produced from ADP and iP Molecular O2 is required for terminal oxidation. PHOTOPHOSPHORYLATION Occurs during photosynthesis inside the chloroplasts. Pigment systemt I & II are involved Sunlight is the external source of energy Molecular Oxygen is not required.

What are rubisco and RuBP And what do they do Briefly list 4 simil.pdf

forecastfashions

Group translocation

AvisekChakraborty6

INTRODUCTION TO THE EVOLUTIONARY METABOLIC MEDICINE BASED ON MITOCHONDRIAL DY...

banafsheh61

Proteinstructuretargetingandsorting 111109072349-phpapp02-1

Bio Slide

Jonathan and Seth

hormone final.pptx

ENERGY-TRANSFORMATION.pdf

AlodiaPastorizo

Bioenergetics [autosaved]

Maira Siddiqui

Photosynthesis

Vikas Kashyap

Protein structure, targeting and sorting

aljeirou

Protein Folding-Self Assembly & Role of Chaperones in Protein Assembly.

Cherry

Similar to Which of the following protons would you NOT expect to find embedded .pdf (20)

1.cell_-_organelles_transport_mechanism.pdf

Chromosome & nuclear structure.pdf

Intracellular trafficking and protein sorting

Cell organelles

Leeuwenhoek described microorganism as Animalcules. Prokaryotes. Euk.pdf

Bio.. mcq..

Role of endoplasmic reticulum in protein systhesis and1

What are rubisco and RuBP And what do they do Briefly list 4 simil.pdf

Group translocation

INTRODUCTION TO THE EVOLUTIONARY METABOLIC MEDICINE BASED ON MITOCHONDRIAL DY...

Proteinstructuretargetingandsorting 111109072349-phpapp02-1

Bio Slide

Jonathan and Seth

hormone final.pptx

ENERGY-TRANSFORMATION.pdf

Bioenergetics [autosaved]

Photosynthesis

Protein structure, targeting and sorting

Protein Folding-Self Assembly & Role of Chaperones in Protein Assembly.

More from SIGMATAX1

Consider a binary search tree T with nodes containing the four fields key, left, right, p. and ht, where key is an integer, left, and right are pointers to the two subtrees, p is a pointer to the parent of the node, and ht is the height of the node (the maximum length of a path from this node to a leaf). Design an algorithm for a procedure insert (T, x) that inserts to a binary searches tree T a new node x (whose key is not in the tree yet), and updates lit fields accordingly. The function should work in O(h) time, where h is the length of the access path. Solution Algorithm is implemented in C: struct node { int key; struct node *left, *right; }; //function to create a new BST node struct node *new_Node(int item) { struct node *tem = (struct node *)malloc(sizeof(struct node)); tem->key = item; tem->left = tem->right = NULL; return tem; } /INSERT PROCEDURE /* Given key in BST,function to insert a new node. */ struct node* insert(struct node* T, int x) { /* If tree is empty then return a new node */ if (T == NULL) return new_Node(x); /* else, repeat down the tree */ if (x < T->x) T->left = insert(T->left, x); else if (key > node->key) T->right = insert(T->right, x); return T; }.

Consider a binary search tree T with nodes containing the four fields.pdf

SIGMATAX1

Based on the article Why It Is Time to Look Beyond Algal Genes in Photosynthetic Slugs By: Cessa Rauch1, Jan de Vries 1, Sophie Rommel2, Laura E. Rose2, Christian Woehle3, Gregor Christa1, Elise M. Laetz4, Heike Wa ¨gele4, Aloysius G.M. Tielens5,6, Jo ¨rg Nickelsen7, Tobias Schumann8, Peter Jahns8, and Sven B. Gould1,* Answer this: --> It has been argued that the reason for detection of such a small amount of mRNA (the read count) of lateral gene transfer origin in the slug cells is due to only very few cells containing plastids. Do you agree with this argument? Why or why not? --> The authors present several alternative reasons for the stability of plastids within the slug cells. Describe two of these reasons. Solution 1. No, we donot agree with the argument. Through LGT from algae to slugs has similar RNA sequence,supported by genome sequence of slug DNA, failed to give an evidence for such LGT events later on. So by this, it is clear that there is no proof that the evolution of kleptoplasty in photosynthetic slugs involves LGT events, where they depend purely on nuclear genes. In some of the photosynthetic slugs, plastids sequestered from algae can continue to photosynthesize for weeks, or even months, in the absence of algal nuclei. Therefore, in conclusion, some isolated algal plastids are by nature more robust than those of land plants and there is no evidence that any of these require the expression of slug genes of algal origin. 2. Plastids have effective photoprotection mechanisms. Slugs acquiring robust plastids will not automatically retain them long-term and endure starvation. Slugs are physiologically adapted and require to retain them functionally..

Based on the articleWhy It Is Time to Look Beyond Algal Genes in P.pdf

SIGMATAX1

Compare/contrast the following classes of hormones by identifying three similarities and four differences among/between them. In addition, be able to list several specific examples of each. AMINE vs. PEPTIDE vs. STEROID Solution Three similarities between the three hormones- 1. Being hormones, the site of synthesis and the site of action of all the three hormones are different. 2. They are all organic molecules and rich in carbon and hydrogen. 3. They are all released directly into the blood stream. Differences Peptide Steroid Amine Composition Made of few to many amino acids(3-200), synthesized from mRNA Derived from cholesterol (from pregnenolone, a steroid precursor) Derived from amino acids tyrosine or tryptophan Solubility in lipid bilayer Not soluble soluble Not soluble Nature Hydrophillic with cell surface receptors( except- thyroid hormones are lipid soluble) Lipophillic with intracellular receptors Hydrophillic with cell surface receptors Mode of Action Through secondary messengers like cAMP, cGMP etc by cascading effect Once inside the cell it binds DNA and modifies transcription Through binding to beta adrenergic receptors, by secondary messengers (signal transduction) or calcium or phosphoinositides Examples ADH, calcitonin, Insulin, Pituitary Hormones (LH,FSH, ACTH, TSH,GH) Estrogen, Progesterone, testosterone, cortisol Epinephrine, nor epinephrine, melatonin, dopamine Peptide Steroid Amine Composition Made of few to many amino acids(3-200), synthesized from mRNA Derived from cholesterol (from pregnenolone, a steroid precursor) Derived from amino acids tyrosine or tryptophan Solubility in lipid bilayer Not soluble soluble Not soluble Nature Hydrophillic with cell surface receptors( except- thyroid hormones are lipid soluble) Lipophillic with intracellular receptors Hydrophillic with cell surface receptors Mode of Action Through secondary messengers like cAMP, cGMP etc by cascading effect Once inside the cell it binds DNA and modifies transcription Through binding to beta adrenergic receptors, by secondary messengers (signal transduction) or calcium or phosphoinositides Examples ADH, calcitonin, Insulin, Pituitary Hormones (LH,FSH, ACTH, TSH,GH) Estrogen, Progesterone, testosterone, cortisol Epinephrine, nor epinephrine, melatonin, dopamine.

Comparecontrast the following classes of hormones by identifying thr.pdf

SIGMATAX1

Why does the incidence of acute disease decrease with age? Solution Incidence of acute diseases decrease with age because the body which is through all the worst times have gained enough resistance though the immunity decreases with age still the body will be ready to fight with acute infections and go over that due to the resistance developed in course of lifetime. i hope this is a brief answer and would help you. Thank you.

Why does the incidence of acute disease decrease with ageSoluti.pdf

SIGMATAX1

White eye color in Drosophila was discovered to be sex linked by Thomas Hunt Morgan, who observed that: White eye color in Drosophila was discovered to be sex linked by Thomas Hunt Morgan, who observed that:A. reciprocal crosses produced different results, which is not true for most autosomal traits.B. the inheritance pattern depended on the sex of the individuals.C. the white- eyed flies were sterile, in contrast to the red-eyed flies.D. A and B.E. All of the above statements apply. Solution The answer is D. A and B. A. reciprocal crosses produced different results, which is not true for most autosomal traits. B. the inheritance pattern depended on the sex of the individuals..

White eye color in Drosophila was discovered to be sex linked by Tho.pdf

SIGMATAX1

Which of these characteristics distinguishes angiosperms from other seed plants? roots Which of these characteristics distinguishes angiosperms from other seed plants?flowersseedstracheids roots Solution 1. Only angiosperms bear true fruit and have flowers. 2.sporophyte and gametophyte generations, swimming sperm and desiccation tolerance 3. monocots (ie cereals) 4. mechanical support (in form of stem and root anchored to the soil) 5. the diploid gametophyte 6. Pollen grains contain the male gametes of gymnosperms, and seeds contain the female gametes called ovules..

Which of these characteristics distinguishes angiosperms from other .pdf

SIGMATAX1

Which of the following is MOST responsible for maintaining the double-stranded structure of DNA? A. Attractive interactions between positively charged purines and negatively charged pyrimidines. B. Hydrogen bonding between the 3\'-OH of deoxyribose and purine and pyrimidine ring nitrogens. C. Hydrophobic interactions between contiguous base pairs and hydrogen bonding between complementary bases on opposite strands. D. Interactions between the phosphate groups in separate chains. E. Hydrophilic interactions between the pentoses in separate chains. Solution A. Attractive interactions between positively charged purines and negatively charged pyrimidines. Explanation- Attractive interactions between partial positive charge on purines and partial negative charge on pyrimidines is responsible for maintaining stability of DNA. Hydrogen bonds in the DNA strands are individually weak but are collectively strong. These bonds between complementary base pairs stabilize the helical structure of DNA. Adenine is paired through two hydrogen bonds with thymine. Guanidine is paired through three hydrogen bonds with cytosine..

Which of the following is MOST responsible for maintaining the doubl.pdf

SIGMATAX1

What are the role of environmental factors on transpiration and how plants defend themselves against water loss? What are the role of environmental factors on transpiration and how plants defend themselves against water loss? Solution Role of environmental factors on transpiration 1. Relative humidity The loss of water vapour from stomata occurs by simple diffusion. When the atmosphere is very humid, the rate of loss of water vapour is lowered. Wind: When the wind velocity is high, rate of transpiration also increases. Temperature: Warm air hold more water vapour so the rate of loss of water increases with increase in temperature. Light: Opening and closing of stomata is regulated by light. More light result in high degree of transpiration. Soil water content: When there is less water content in the soil , the leaf droops / get dehydrated making the stomata to close resulting in less rate of transpiration. Carbon dioxide concentration When the concentration is high, stomata closes resulting in low rate of transpiration and vice versa. Plants defense mechanism against water loss Plants require water for it\'s survival. So it will have to regulate water loss ( especially during drought) . Several leaf adaptations help to combat water loss. Leaf arrangements play important role, transversely arranged leaves loss more water than perpendicularly arranged ones. Other anatomical features of the leaves such as sunken/vestigial stomata and closely arranged mesophyll cells reduces transpiration. Presence of hair, cuticle(waxy) , gums, mucilage etc on leaf surface help reducing transpiration. During drought, plant hormone abscisic acid play an important role by closing the stomata reducing water loss..

What are the role of environmental factors on transpiration and h.pdf

SIGMATAX1

What will be quantization step size in numbers and in voltage for th.pdf

SIGMATAX1

What is the need for a shadow page table? (One or two brief sentences please) Solution Virtualization allows multiple operating system instances to run concurrently on a single computer. Virtualization is the creation of a virtual rather than actual version of something, such as an operating system, a server, a storage device or network resources. Virtualization is where the framework divides the resource into one or more execution environments Even something as simple as partitioning a hard drive is considered virtualization because you take one drive and partition it to create two separate hard drives. Types of virtualization: Storage virtualization: is the pooling of physical storage from multiple network storage devices into what appears to be a single storage device Data virtualization: is abstracting the traditional technical details of data and data management, such as location, performance or format.. Network virtualization: is a method of combining the available resources in a network by splitting up the available bandwidth into channels, each of which is independent from the others. Server virtualization: the partitioning a physical server into smaller virtual servers. Shadow Tables are objects used to improve the way machines, networks and programs handle information. Shadow page tables are used by the hypervisor to keep track of the state in which the guest thinks its page tables should be. The guest can\'t be allowed access to the hardware page tables because then it would essentially have control of the machine. So, the hypervisor keeps the \"real\" mappings guest virtual -> host physical in the hardware when the relevant guest is executing, and keeps a representation of the page table..

What is the need for a shadow page table (One or two brief sentences.pdf

SIGMATAX1

What step of protein synthesis is shown in the figure? elongation stage of transcription termination stage of translation mRNA processing elongation stage of translation initiation stage of transcription Solution elongation stage of transcription Explanation: newly synthesized mRNA separates from the DNA template as the RNA plymerase continues elongation. multiple RNA plymerases transcribe simultaneously..

What step of protein synthesis is shown in the figure elongation st.pdf

SIGMATAX1

Water is forced through a contraction causing low pressure. The wat.pdf

SIGMATAX1

Using a PIC 18 Write a time overflow interrupt routine that detects .pdf

SIGMATAX1

The three types of mixtures are and The weak bond forming a bridge between a hydrogen atom and nitrogen or oxygen is a bond. A is an inorganic compound which has a bitter taste, feels slipper and is a proton acceptor. Solution 10. Three types of mixtures are suspensions, solutions and colloids 11.answer is hydrogen bond. A hydrogen bond is the electrostatic attraction between two polar groups that occurs when a hydrogen (H) atom covalently bound to a highly electronegative atom such as nitrogen (N), oxygen (O), or fluorine (F) experiences the electrostatic field of another highly electronegative atom nearby. 12.answer is Base..

The three types of mixtures are and The weak bond forming a bridge .pdf

SIGMATAX1

Some commercial airplanes recirculate approximately 50% of the cabin air in order to increase fuel efficiency. The researchers studied 1096 airline passengers, among which some traveled on airplanes that recirculated air and others traveled on planes that did not recirculate air. Of the 515 passengers who flew on planes that did not recirculate air, 106 reported post-flight respiratory symptoms, while 112 of the 581 passengers on planes that did recirculate air reported such symptoms. Is there sufficient evidence to conclude that the proportion of passengers with post- flight respiratory symptoms differs for planes that do and do not recirculate air? Test the appropriate hypotheses using a = 0.05. You may assume that it is reasonable to regard these two samples as being independently selected and as representative of the two populations of interest. (Use a statistical computer package to calculate the P-value. Use Pdo not recirculate - Pdo recirculate- Round your test statistic to two decimal places and your P-value to four decimal places.) Solution Null , There Is No Significance between them Ho: p1 = p2 Alternate , There Is Significance between them H1: p1 != p2 Test Statistic Sample 1 : X1 =106, n1 =515, P1= X1/n1=0.206 Sample 2 : X2 =112, n2 =581, P2= X2/n2=0.193 Finding a P^ value For Proportion P^=(X1 + X2 ) / (n1+n2) P^=0.199 Q^ Value For Proportion= 1-P^=0.801 we use Test Statistic (Z) = (P1-P2)/(P^Q^(1/n1+1/n2)) Zo =(0.206-0.193)/Sqrt((0.199*0.801(1/515+1/581)) Zo =0.54 | Zo | =0.54 Critical Value The Value of |Z | at LOS 0.05% is 1.96 We got |Zo| =0.54 & | Z | =1.96 Make Decision Hence Value of |Zo | < | Z | and Here we Do not Reject Ho P-Value: Two Tailed ( double the one tail ) -Ha : ( P != 0.5403 ) = 0.589 Hence Value of P0.05 < 0.589,Here We Do not Reject Ho No sufficient evidence to condude that the proportion of passengers with post-flight respratory symptoms differs for planes that do and do not recirculate air ANS: Z = 0.54 P = 0.589.

Some commercial airplanes recirculate approximately 50 of the cabin .pdf

SIGMATAX1

Solve for 02 sin 2+sin 5=0. Separate your answers with commas.pdf

SIGMATAX1

Required: 1. Briefly describe different computer controls available to an organization. a. Identify how these controls help to maintain system security and reduce risk. b. Do companies need a formal internal policy that addresses security violations? Why/why not? c. Consider what type of actions a company should take against employees that circumvent computer Solution Security Controls: Administrative Controls, Technical or Logical Controls, Physical Controls The constituents of Security: Vulnerability, Threat, Risk, Exposure, Countermeasure or Safeguard The data vendor grants access needs or may choose to give this function to business unit managers. And it is the data owner who will contract with security violations pertaining to the data he is liable for defending. Subjects that should be measured when developing a security policy: Classification Controls: Qualitative Risk Analysis Techniques:.

Required 1. Briefly describe different computer controls available .pdf

SIGMATAX1

a. What is Snort? [1] b. According to “Snort Users Manual” (downloadable under “Docs” menu), what are the three modes in which Snort operates and how it behaves under each mode? Show an example command to put Snort in each mode. [6] c. Explain the difference between the rule headers and rule options [2] d. List 8 different types of actions Snort may take upon a rule match. Describe what each action type does. [2] Solution a. Snort: Snort is an open source network capable of performing real-time traffic study and packet-logging on IP networks. It can transmit the protocol analysis, content searching & matching and is used to detect a range of attacks and investigates, such as buffer run over, cautious port scans, OS fingerprinting efforts. d. Types of snort actions:.

a. What is Snort [1]b. According to “Snort Users Manual” (downloa.pdf

SIGMATAX1

Proponents of decentralized forms of socialism usually supportA co.pdf

SIGMATAX1

Please convert the following C code to assembly Y86 int i,j; ..... if (i >j) { i= i+5; } else { i=0; j++; } Solution Answer: MIPS Assembly Language Code : .zero 1 main: push rbp mov rbp, rsp mov eax, DWORD PTR [rbp-4] cmp eax, DWORD PTR [rbp-8] jle .L2 add DWORD PTR [rbp-4], 5 jmp .L3 .L2: mov DWORD PTR [rbp-4], 0 add DWORD PTR [rbp-8], 1 .L3: mov eax, 0 pop rbp ret __static_initialization_and_destruction_0(int, int): push rbp mov rbp, rsp sub rsp, 16 mov DWORD PTR [rbp-4], edi mov DWORD PTR [rbp-8], esi cmp DWORD PTR [rbp-4], 1 jne .L7 cmp DWORD PTR [rbp-8], 65535 jne .L7 mov edi, OFFSET FLAT:std::__ioinit call std::ios_base::Init::Init() mov edx, OFFSET FLAT:__dso_handle mov esi, OFFSET FLAT:std::__ioinit mov edi, OFFSET FLAT:std::ios_base::Init::~Init() call __cxa_atexit .L7: nop leave ret push rbp mov rbp, rsp mov esi, 65535 mov edi, 1 call __static_initialization_and_destruction_0(int, int) pop rbp ret.

Please convert the following C code to assembly Y86int i,j; ......pdf

SIGMATAX1

More from SIGMATAX1 (20)