Securing the data in a database is typically as important a concern .pdf

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Securing the data in a database is typically as important a concern as is its design. Research database security on the web, and then respond to the following: Describe a few security risks that databases must be protected against. Describe some steps that can be taken to ensure that a database is secured. Solution Risks : 1) excess privileges : many users are provided with default priveleges out of which some are not necessary for them and may be misused. 2) database injection attacks like SQL and No-SQL attacks are used to fetch confidential data and manipulate it 3)malware - application that capture private data 4) Unprotected storage media like backup disks and tapes. 5) lack of professional to handle security mechanisms. Mitigation steps : 1)monitoring DB access activity with user identification 2) remove excess rights 3) train more number of secirity professionals 4) block malicious requests to data.

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Cloud Storage is a branch of Cloud Computing, which plays an important role in IT world. Cloud providers are providing a huge volume of storage space as per the user needs. Due to wide usage of this, it also increases data security issues and threats. Hence efforts are being made to encrypt the data stored in the cloud. In this paper, we are going to look at different encryption and auditing techniques that are used to avoid data breaching in cloud storage. Nikhil Sreenivasan ""Data Storage Issues in Cloud Computing"" Published in International Journal of Trend in Scientific Research and Development (ijtsrd), ISSN: 2456-6470, Volume-4 | Issue-2 , February 2020, URL: https://www.ijtsrd.com/papers/ijtsrd30194.pdf Paper Url : https://www.ijtsrd.com/computer-science/computer-network/30194/data-storage-issues-in-cloud-computing/nikhil-sreenivasan

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Overview To Database Security. What is Database Security Why need of database security. Concepts of Database Security. Security Problems Security Controls In today’s world, we need everything secured whether it is your mobile phone , computer , vehicle or almost anything. What is database security? Database It is a collection of information stored in a computer. Security: It is being free from danger. Database Security: It is the mechanisms that protect the database against intentional or accidental threats. Database Security is defined as the process by which “Confidentiality, Integrity and Availability” of the database can be protected Why need of database security?If there is no security to database what happens??? Data will be easily corrupted It is important to restrict access to the database from authorized users to protect sensitive data. Concepts of Database SecurityThree are 3 main aspects Secrecy or Confidentiality Integrity Availability SECRECY /It is protecting the database from unauthorized users. Ensures that users are allowed to do the things they are trying to do. Encryption is a technique or a process by which the data is encoded in such a way that only that authorized users are able to read the data. INTEGRITYProtecting the database from authorized users. Ensures that what users are trying to do is correct. For examples, An employee should be able to modify his or her own information. AVAILABILITYDatabase must have not unplanned downtime. To ensure this ,following steps should be taken Restrict the amount of the storage space given to each user in the database. Limit the number of concurrent sessions made available to each database user. Back up the data at periodic intervals to ensure data recovery in case of application users.

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Database security is a growing concern as the amount of sensitive data collected and retained in databases is fast growing and most of these data are being made accessible via the internet. Majority of the companies, organizations and teaching and learning institutions store sensitive data in databases .As most of these data are electronically accessed , It can therefore be assumed that , the integrity of these numerous and sensitive data is prone to different kind of threat such as{Unauthorized access, theft as well access denial}. Therefore, the need for securing databases has also increased The primary objectives of database security are to prevent unauthorized access to data, prevent unauthorized tampering or modification of data, and to also ensure that, these data remains available whenever needed. In this paper, we developed a database security framework by combining different security mechanism on a sensitive students information database application designed for Shehu Shagari College of Education Sokoto (SSCOE) with the aim of minimizing and preventing the data from Confidentiality, Integrity and Availability threats

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It was noted that excess water can kill plants. If water is not toxic, and if all plants require water, then how can excess water kill plants? What is actually killing the plants? Solution When excess water is present, the roots will be deprived of oxygen because the water fills the pores in the soil. This mostly occurs in clay soils. Also, excess water drain out the nutrients and promote root rot disease, which gradually kills the plant..

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In biology, the roles of transition metal ions can be broadly grouped into two classifications: redox chemistry (catalysis, electron transfer, etc.) and acid-base chemistry (catalysis, structural, etc.). A) List three metal ions that commonly participate in biological redox chemistry, and give examples of how they participate. B) Compare this list to metal ions commonly participating in acid-base chemistry. Discuss the overlap of these two groups of metal ions. (e.g. Why are some used for one of these functions and not for the other?) C) List some common biological ligands of these metal ions. Can you see patterns that would allow you to group these ligands into sets with common properties, or that bind common metals? D) Why are trace metals used for these biological functions, and not other more-abundant elements like C,N,O, etc? Solution (a) iron, copper, cobalt, manganese etc are used in biology redox reactions like molecular di oxygen, superoxide, hydrogen peroxide thus inhibiting the effect or formation of these compound by use of the certain other enzymes.. (c) the ions and the Ligands for them are Ike gluthaione peroxide which can bind to chromium, arsenic etc.. Iron binds to ferrodoxin which is associated with the photosynthesis.. (d) transition metals are used because they have empty d-orbitals in with them due to which they can easily transfer or accept the electrons or exchange the electrons among the metals and enzymes.. Which is not possible in other trace metals like carbon, nitrogen etc...

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Explain how recombination increases the amount of genetic variation in offspring: Explain why it is not possible to have a recombination frequency of greater than 50% (half recombinant progeny): A second pair of Drosophila are mated. The female is Cucu YY (straight wing, gray body), while the male is Cucu yy (straight wing, yellow body). Assuming recombination, perform the cross and list the offspring genotypes and phenotypes. Solution Que-3: The recombination frequency is the measure of genetic linkage at different loci. The value of recombination frequency is 50% if the chromosomal genes are located at different regions on the different chromosome because of independent assortment. If the genes are located closely together on the same chromosome, then those genes considered as genetically linked; because they have not assorted independently due to \"crossing over\", then the recombination frequency value is less than 50%. Linked genes are due to linkage often do not obey Mendel\'s laws of inheritance such as independent assortment because of \"crossing over with reciprocal exchange of hereditary genetic material meticulously in between non-sister chromatids\" at the time of \"synapsis formation\" (propahse -I) in reduction division or meiosis 1. These events are leading to formation of gamete cells with complete different genotypically from diploid parent reproductive cell via meiosis finally induce \"genetic variation\" in phenotype due to \"fertilization\" events between sperm cells & ovum. The recombination of chromosomes enables meiosis process of cell division with the generation of novel recombinant nucleotide -sequences by crossing over during the cell division in both unicellular and multicellular species. Sometimes, double strand breaks in DNA during homologous chromosomal recombination may be produced due to exposure any harmful radiation finally may cause higher genetic variation via duplication followed by fertilization to form a phenotype with higher genetic variation Que-4: Crossing over is somewhat randomly distributed over the length of the chromosome because if the two genes are close together then it is very change to get exchange of chromosomal breaks therefore, if the two genes are located farther apart then there will be higher room for the exchange of genetic material with chromosomal breaks. Therefore, when two genetic are far apart on the same chromosome are more likely to have a crossover between them than two loci that are close together. This recombination frequency is considered as a measure of genetic linkage during the crossover of the homologous chromosomes. Linkage disequilibrium defined as the existence of alleles at different loci with the absence of genetic linkage between them even though there is no equilibrium with allelic frequencies independently. The recombination frequency is the measure of genetic linkage at different loci. The value of recombination frequency is 50% \"if the chromosomal genes are .

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8. List 4 GENERAL ways that a CELL, can DECREASE the rate of its reactions? 1. 2. 3. 4. 10. What is the BASIC difference between substrate-level phosphorylation and oxidative phosphorylation? (do not tell me which pathways are involved). 11. Exactly….and I mean exactly…..which chemical reaction in the aerobic metabolic pathways, actually uses oxygen, and what does it do to the oxygen that it uses. Solution 8). Enzymes are the proteins, which decrease the activation energy of reaction and catalyse the reaction process. All enzymes are catalysts because, Cells decrease the rate of reaction by 1). Altering pH 2). Accumulation of products 3). Altered temperature 4). Inactivation of catalyst.

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Among 10 people traveling in a group, 2 have outdated passports. It is known that inspectors will check the passports of 20% of the people in any group passing their desks. The group can go as a whole (all 10) to one desk or can split into two groups of 5 and use two different desks. How should members of the group arrange themselves to maximize the probability of getting by the inspectors without having the outdated passports detected? I do not know how to figure out each probability... The section is on counting rules - i am guessing it is without replacement and order does not matter. Thank you for your help Solution Inspectors would check 20% of people in the group Hence if 10 people, 2 would be checked or if 5 one person would be checked If 10 people go to one desk, p for having outdated passport = 0.2 X - no of people having correct passports in 10 is binomial with p = 0.8 They would be caught only if Inspector selects any one person from the two. Y - no of people having outdated passports and p for Y =0.2 P(Y>=1) = 1-P(Y=0) = 1-(0.8)10 = 0.8926 --------------------------------------------- If goes into two groups then either each group can have one outdated or one group no outdated and other groups 2 outdated. For each group one outdated prob of being caught = P(y=1) in I group + P(y=1 in second group)-P( P(y=1) in I group *P(y=1 in second group) as these two events are independent = {1-(0.8)5}+{1-(0.8)5}-{1-(0.8)5}{1-(0.8)5} =2(0,67232)-0.672322 = 0.8926 (same as the first case) ---------------------------------------------- If one group has both outdated and other has none then prob of being not caught = 1(Prob of atleast one in the group having outdated 2) Here p for good passport =0.6 = P(Y>=1) = 1-P(y=0) = 1-(0.6)5 =0.92224 Hence this is the best method to go. Divide into two groups with 2 outdated in one group.

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1.) The normal eye color of Drosophila is red, but strains in which all flies have brown eyes are available. Similarly, wings are normally long, but there are strains with short wings. A female from a pure line with brown eyes and short wings is crossed with a male from a normal pure line. The F1 consists of normal females and short-winged males. An F2 is then produced by intercrossing the F1. Both sexes of F2 flies show phenotypes as follows: 3/8 red eyes, long wings 3/8 red eyes, short wings 1/8 brown eyes, long wings 1/8 brown eyes, short wings Deduce the inheritance of these phenotypes; use clearly defined genetic symbols of your own invention. State the genotypes of all three generations and the genotypic proportions of the F1 and F2. 2.) When a cell of genotype A/a ; B/b ; C/c having all the genes on separate chromosome pairs divides mitotically, what are the genotypes of the daughter cells? 3.) A presumed dihybrid in Drosophila, B/b ; F/f, is test-crossed with b/b ; f/f. (B = black body ; b = brown body; F = forked bristles; f = unforked bristles.) The results are black, forked 230 brown, forked 240 black, unforked 210 brown, unforked 250 Use the 2 test to determine if these results fit the results expected from testcrossing the hypothesized dihybrid. 4.) In mice, dwarfism is caused by an X-linked recessive allele, and pink coat is caused by an autosomal dominant allele (coats are normally brownish). If a dwarf female from a pure line is crossed with a pink male from a pure line, what will be the phenotypic ratios in the F1 and F2 in each sex? (Invent and define your own gene symbols.) Solution Answer 2. If a cell divides mitotically (equational division), the genetic constitution of the daughter cells remains the same as the parents. So a parent with the genotype A/a; B/b; C/c will give rise to daughter cells with the genotype A/a; B/b; C/c. However, if the cell was to divide meiotically (reductional division), the genetic constitution would be different as only one half of the parental chromosomes goes to the daughter cell. So the genotypes would be as follows: ABC or AbC or ABc or Abc or aBC or abC or aBc or abc Answer 3. For the given cross BbFf X bbff For such a test cross the resultant phenotypic ration obtained is 1 (BF):1 (Bf):1 (bF):1 (bf). For dihybrid cross, the degree of freedom will be 4 – 1 = 3 Chi square test is as follows BF Bf bF bf Totals Observed phenotype 230 210 240 250 930 Expected phenotype 930x ¼= 232.5 930x ¼= 232.5 930x ¼= 232.5 930x ¼= 232.5 930 Observed – expected -2.5 -22.5 7.5 17.5 (Obs – Exp)2 Expected 6.25/232.5= 0.0268 506.25/232.5= 2.1774 56.25/232.5= 0.2419 306.25/232.5= 1.3172 3.7633 Looking at the probability value table for degree of freedom = 3, df .90 .70 .50 .30 .20 .10 .05 .01 3 0.58 1.42 2.37 3.67 4.64 6.25 7.82 11.35 Hence we can deduce that there is more than 20% probability for the occurrence of this event. Answer 4. Let XD = Normal phenotype (tall) Xd = Dwarf phenotype P = Pink coat p = Br.

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29. Code an application program that keeps track of student information at your college. Include their names, identification numbers, and grade point averages in a fully encapsulated, homogeneous sorted singly linked list structure. When launched, the user will be asked to input the initial number of students and the initial data set. Once this is complete, the user will be presented with the following menu: Enter: 1 to insert a new student\'s information, 2 to fetch and output a student\'s information, 3 to delete a student\'s information, 4 to update a student\'s information, 5 to output all the student information in sorted order, and 6 to exit the program. The program should perform an unlimited number of operations until the user enters a 6 to exit the program. If the user requests an operation on a node not in the structure, the program output should be “node not in structure.” Otherwise, the message “operation complete” should be output. in java Solution Dear Asker, Following 3 classes consist the colution to this problem. This will give you an idea to approach this problem: //Student Class package studentinfo; public class Student implements Comparable { private String name; private String id; private double gpa; public Student(String id, String name, double gpa) { this.name = name; this.id = id; this.gpa = gpa; } /** * @return the name */ public String getName() { return name; } /** * @param name the name to set */ public void setName(String name) { this.name = name; } /** * @return the id */ public String getId() { return id; } /** * @param id the id to set */ public void setId(String id) { this.id = id; } /** * @return the gpa */ public double getGpa() { return gpa; } /** * @param gpa the gpa to set */ public void setGpa(double gpa) { this.gpa = gpa; } @Override public int compareTo(Student o) { if (this.id.compareTo(o.id) < 0) { return -1; } else if (this.id.compareTo(o.id) > 0) { return 1; } else if (this.gpa < o.gpa) { return -1; } else if (this.gpa > o.gpa) { return 1; } return 0; } public boolean equals(Student o) { if (this.name == o.name && this.id == o.id && this.gpa == o.gpa) { return true; } return false; } } //Student container class package studentinfo; import java.util.Collections; import java.util.LinkedList; public class StudentContainer { LinkedList studentList; public StudentContainer() { this.studentList = new LinkedList(); } public boolean hasStudent(String id, String name, double gpa) { return this.studentList.contains(new Student(id, name, gpa)); } public boolean insertStudent(String id, String name, double gpa) { if (!this.isValidString(id) || !this.isValidString(name) || !this.isValidGpa(gpa)) { return false; } this.studentList.add(new Student(id, name, gpa)); return true; } public boolean deleteStudentById(String id) { Student toRemove = null; for (Student e : studentList) { if (e.getId() == id) { toRemove = e; break; } } if (toRemove != null) { this.studentList.remove(toRemove); return true; } return false;.

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Write the code for a small function called myStack, which creates a Stack object of String elements, named cities. Add \"Sydney\", then \"London\", then \"New York\" and then \"Beijing\" to cities. Then print each element on the screen, one per line. Print each element in a way that causes cities to be empty (i.e. have 0 elements) afterwards. (6 mks) Solution PROGRAM CODE: package stack; import java.util.Stack; public class MyStackClass { //stack function public static void myStack() { //creating a stack object called cities Stack cities = new Stack(); //Adding cities into the stack cities.push(\"Sydney\"); cities.push(\"London\"); cities.push(\"New York\"); cities.push(\"Beijing\"); //using while loop to loop through the stack till its empty while(!cities.isEmpty()) { // pop function is used to remove elements from the stack // pop() returns the elemets which is printed on the csreen System.out.println(cities.pop()); } } // Main function for the class public static void main(String args[]) { myStack(); // function call to the stack method } } OUTPUT: Beijing New York London Sydney.

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Why does the neutralization of an acid by a base often produce water? Solution Neutralisation refers to a reaction between an acid and a Base. This reaction often results in the production of water. It is due to a fact that an acid liberates H+ ions while as a Base releases OH- ions. These ions combine to form water. For example: HCl(aq) H+(aq) + Cl(aq) NaOH(aq) Na+(aq) + OH(aq) HCl + NaOH NaCl + H2O..

Why does the neutralization of an acid by a base often produce water.pdf

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Which statement about phloem transport is false? It takes place in sieve tubes It requires metabolic energy A high pressure potential is maintained in the sieve tubes At sinks, solutes are actively transported into sieve tube elements The mycorrhizal relationship allows a plant’s roots to ?? increase the amount of soil than can be accessed for nutrients. grow in nutrient-poor soils. avoid direct interaction with the soil. None of the above Solution 1. At sinks, solutes are actively transported into sieve tube elements. (False) Phloem is a vascular tissue in plants that actively transports synthesized food from the source tissue ( photosynthetic) to sink tissue (non-photosynthetic). The sieve tube elements (STEs) are the phloem tissue compartments which help in the transportation of food from source to sink. Phloem transport requires metabolic energy and a high-pressure potential is maintained at the sieve tubes. 2. Increase the amount of soil that can be accessed for nutrients Mycorrhiza is the symbiotic relationship between fungi and the plant roots on which they extensively grow. The network of fungal growth or mycorrhiza helps in increasing the surface area of the roots so that they can access more amount of soil and take up nutrients essential for plant growth..

Which statement about phloem transport is falseIt takes place in .pdf

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