In the Meselson Stahl experiment, E. coli was grown for many generati.pdf

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In the Meselson Stahl experiment, E. coli was grown for many generations in a medium \"heavy\" nitrogen^15 N. Once the bacterial DNA was uniformly labeled, the bacteria were transferred to a medium containing the \"light\" nitrogen, ^14N. After ONE round of replication, DNA was extracted and subjected to density gradient centrifugation in cesium chloride. Upon inspection, there was only a single band of intermediate density DNA in the gradient. This proved that DNA replication CANNOT be conservative dispersive semiconservative conservative of dispersive continuous. Solution Answer is Dispersive replication. Dispersive replication would result in double-stranded DNA with both strands having mixtures of 15N and 14N DNA, either of which would have appeared as DNA of an intermediate density..

In the Meselson Stahl experiment, E. coli was grown for many generations in a medium
"heavy" nitrogen^15 N. Once the bacterial DNA was uniformly labeled, the bacteria were
transferred to a medium containing the "light" nitrogen, ^14N. After ONE round of replication,
DNA was extracted and subjected to density gradient centrifugation in cesium chloride. Upon
inspection, there was only a single band of intermediate density DNA in the gradient. This
proved that DNA replication CANNOT be conservative dispersive semiconservative
conservative of dispersive continuous.
Solution
Answer is Dispersive replication.
Dispersive replication would result in double-stranded DNA with both strands having mixtures
of 15N and 14N DNA, either of which would have appeared as DNA of an intermediate density.

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questions. How did semiconservative? and Stahl s experiments support the idea that DNA replication is 2. A new form of DNA is discovered that appears to be able to replicate itself both in the 3\' 5\' direct and in the sequence direction. If this is true, how would newly on replication differ from DNA replication as we this discovered DNA know it? 3. Amazingly, an alien species of cellular organism is found alive in the remains of a meteorite that landed in the Mojave Desert. As a scientist, you are trying to determine whether this alien life-form uses DNA, protein, or some other type of compound as its hereditary material a. What kinds of experiments would you propose to determine what the hereditary material b. If the hereditary material tums out to be similar to our DNA, describe the simplest experiments you could run to try to determine if is double-stranded like our DNA, triple-stranded, or some other configuration. Copyright 2014 Pearson Activity 15.3 Inc. Solution Answer to question 1 Nitrogen is one of the major constituent of DNA. 14N is by far the most abundant isotope of nitrogen, but DNA with the heavier 15N isotope is also functional. E. coli was grown for several generations in a medium containing NH4Cl with 15N. When DNA is extracted from these cells and centrifuged on a salt density gradient, the DNA separates out at the point at which its density equals that of the salt solution. The DNA of the cells grown in 15N medium had a higher density than cells grown in normal 14N medium. After that, E. coli cells with only 15N in their DNA were transferred to a 14N medium and were allowed to divide. DNA was extracted periodically and was compared to pure 14N DNA and 15N DNA. After one replication, the DNA was found to have intermediate density. Since conservative replication would result in equal amounts of DNA of the higher and lower densities (but no DNA of an intermediate density), conservative replication was excluded. However, this result was consistent with both semiconservative and dispersive replication. Semiconservative replication would result in double-stranded DNA with one strand of 15N DNA, and one of 14N DNA, while dispersive replication would result in double-stranded DNA with both strands having mixtures of 15N and 14N DNA, either of which would have appeared as DNA of an intermediate density. Meselson–Stahl continued to sample cells as replication continued. DNA from cells after two replications had been completed was found to consist of equal amounts of DNA with two different densities, one corresponding to the intermediate density of DNA of cells grown for only one division in 14N medium, the other corresponding to DNA from cells grown exclusively in 14N medium. This was inconsistent with dispersive replication, which would have resulted in a single density, lower than the intermediate density of the one-generation cells, but still higher than cells grown only in 14N DNA medium, as the original 15N DNA would have be.

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In what ways do the experts foresee the use of both virtualization and cloud computing evolving in the future? Solution Use ofVirtualization and cloud computing security : • Self-defending VM security • Layered coordinated defenses • Security optimized for virtual and cloud environments • Visibility, reporting, and auditing • Encryption for virtual and cloud environments • Security that travels with data Security solutions should offer both agent-less and agent-based security options to provide flexible deployment alternatives and close security gaps unique to virtualized and cloud environments. Encryption addresses a range of security challenges related to virtualization and cloud computing. Standard 128-, 192-, or 256-bit encryption of storage volumes deters hackers from prying and thieving, and reduces the risk that the cloud storage devices could be sold or reused while they still contain confidential or private information. Encryption also greatly reduces the risk of malicious VM attacks; as long as the encryption key for the data stores have not been provided, even if rogue VMs reach data stores, volumes are unmountable and unreadable. Encryption with enterprise-controlled key management enables IT to comply with security best practices, internal governance, and external regulation. Data is kept secure, and the key management solution can provide monitoring, reporting, and auditing capabilities that provide visibility into data access. As a result, enterprises realize a significant reduction in the scope of compliance audits. Encryption for virtual and cloud environments: In summary, best practices for encryption for virtual and cloud data protection include: • Integration with leading cloud service providers and virtual environments. • Policy-based key management that determines where and when encrypted data can be accessed. • Identity-based and integrity-based server validation to determine which servers can access secure storage volumes and whether security is up-to-date on those servers prior to data access. • Business control of encryption keys, either on-site or through a separate SaaS service, to maintain a strict separation of duties between the business and cloud service provider..

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implement the ListLinked ADT (the declaration is given in ListLinked.h)(60 points) - implement the following operations: - constructor, assignment operator, destructor - insert, remove, replace, clear - isFull, isEmpty - gotoBeginning, gotoEnd, gotoNext, gotoPrior, getCursor implement the function moveToBeginning() that removes the data item marked by the cursor and reinserts it at the beginning of the list - implement the function insertBefore(..) that will insert the new data item before the cursor or if the list is empty as the first element of the list #include \"ListLinked.h\" // ListNode member functions template List::ListNode::ListNode(const DataType& nodeData, ListNode* nextPtr) { this->dataItem = nodeData; this->next = nextPtr; } // List member functions template List::List(int ignored = 0) { } template List::List(const List& other) { } template List& List::operator=(const List& other) { } template List::~List() { } template void List::insert(const DataType& newDataItem) throw (logic_error) { } template void List::remove() throw (logic_error) { } template void List::replace(const DataType& newDataItem) throw (logic_error) { } template void List::clear() { } template bool List::isEmpty() const { return false; } template bool List::isFull() const { return false; } template void List::gotoBeginning() throw (logic_error) { } template void List::gotoEnd() throw (logic_error) { } template bool List::gotoNext() throw (logic_error) { return false; } template bool List::gotoPrior() throw (logic_error) { return false; } template DataType List::getCursor() const throw (logic_error) { DataType t; return t; } template void List::moveToBeginning () throw (logic_error) { } template void List::insertBefore(const DataType& newDataItem) throw (logic_error) { } #include \"show5.cpp\" Solution #include #include #include using namespace std; struct Node { int data; Node *next; }; Node *tail=NULL; Node* Insert_atHead(Node *head, int data) { Node *temp=new Node; temp->data=data; temp->next=head; head=temp; return head; } Node* Insert_atTail(Node *head, int data) { Node *tail=head; if(tail==NULL) { Node *temp = new Node; temp->data=data; temp->next=NULL; head=temp; return head; } while(tail!= NULL) { if(tail->next==NULL) { Node *temp=new Node; temp->data=data; temp->next=NULL; tail->next=temp; return tail; } else { tail=tail->next; } } } Node *Insert_atPosition(Node *head, int data, int position) { int i=0; int j=position-1; if(position==0) { Node *temp = new Node; temp->data=data; temp->next=head; head=temp; return head; } else { Node *temp=head; while(inext; i++; } Node *temp1 = new Node; temp1->data=data; temp1->next=temp->next; temp->next=temp1; return head; } return head; } Node* Delete(Node *head, int position) { Node *temp=head; int i=1; int j=position; if(j==0) { head=head->next; return head; } Node *temp1 = new Node; while(j>0) { temp1=temp; temp=temp->next; j--; } temp1->next=temp->next; delete temp; return head; } int GetNode(Node *head,int posi.

implement the ListLinked ADT (the declaration is given in ListLinked.pdf

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