Problem I. As shown in the pedigree below, a pair of first cousins (.pdf

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Problem I. As shown in the pedigree below, a pair of first cousins (I and K) marry an unrelated pair of first cousins (J and L). Their children M and N produce child O. Q1. Calculate the inbreeding coefficient (f) of O (express answers to Q1 and Q3 to three decimal places). Assume that f= 0 for all ancestors. (Hint: To help you out, I have shaded individuals who cannot contribute to autozygosity.) To show your work, use the letter names to indicate each path/loop that contributes to autozygosity. For example: path 1 = M-I-E… Q2. Interpret your result. What does this value of f indicate? Q3. Recalculate f for child O when f=1/2 for great-great grandmother D (f=0 for all other ancestors). Hint: After you have calculated all the loops above, you’ll adjust just the loop that includes ancester D to account for her ancestral inbreeding using the formula presented in class (and text). B 0271 M H K E Solution Path 1: · O-M-I-E-A-F-K-N-O · O-M-I-E-B-F-K-N-O So, the probability that a copy of gene in A makes the way down to O via E and a copy of gene in B makes the way down to O via F =(½)n = (½)7=0.00781. Here n= node=7 Path 2: · O-N-L-H-D-G-J-M-O · O-N-L-H-C-G-J-M-O Similarly, like the first one the total probability in path 2 will be same that is 0.00781. Thus inbreeding co-efficient = 0.00781+0.00781=0.01562. 2. Generally, the higher the level of inbreeding closer the coefficient of relationship approaches a value of 1 and approaches a value of 0 for individuals with remote common ancestors. As here the value of inbreeding co-efficient is approximately 0 so we can conclude that 1. the common ancestors are not inbred. 2. O has remote common ancestors means they are distantly related. 3. When a common ancestor is itself inbred the inbreeding coefficient is = (1/2)n (1+ inbreeding co-efficient of common ancestor) , The probability that a copy of gene in D makes the way to O = (1/2)7 = 0.00781 Therefore, inbreeding co-efficient is = 0.00781 * (1 + 1/2) =0.01172.

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Predefined Characters in C Type. Please make sure the program compiles thanks :) // This program reads one character from the keyboard and will // - convert it to uppercase, if it is lowercase // - convert it to lowercase, if it is uppercase // - display a message, if it is not a letter of the alphabet #include #include using namespace std; int main() char c coutc if (isalpha(c)) //check to see if it is a letter of alphabet if (isupper(c))//check to see if it is uppercase cout Solution #include #include using namespace std; int main() { char c; cin.get(c); //cin.get() input the character which can be of any type if(isalpha(c)) { if(isupper(c)) { cout<<\"Your character \"<.

Predefined Characters in C Type. Please make sure the program compil.pdf

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Objective: Binary Search Tree traversal (2 points) Use traversal.pptx as guidance to write a program to build a binary search tree Dictionary. Input records from inventory.txt. Both key and Element of BST have the same data from each input record. public interface BinNode { /** Get and set the element value */ public E element(); public void setElement(E v); /** @return The left child */ public BinNode left(); /** @return The right child */ public BinNode right(); /** @return True if a leaf node, false otherwise */ public boolean isLeaf(); } import java.lang.Comparable; /** Binary Search Tree implementation for Dictionary ADT */ class BST, E> implements Dictionary { private BSTNode root; // Root of the BST int nodecount; // Number of nodes in the BST /** Constructor */ BST() { root = null; nodecount = 0; } /** Reinitialize tree */ public void clear() { root = null; nodecount = 0; } /** Insert a record into the tree. @param k Key value of the record. @param e The record to insert. */ public void insert(Key k, E e) { root = inserthelp(root, k, e); nodecount++; } // Return root public BSTNode getRoot() { return root; } /** Remove a record from the tree. @param k Key value of record to remove. @return The record removed, null if there is none. */ public E remove(Key k) { E temp = findhelp(root, k); // First find it if (temp != null) { root = removehelp(root, k); // Now remove it nodecount--; } return temp; } /** Remove and return the root node from the dictionary. @return The record removed, null if tree is empty. */ public E removeAny() { if (root == null) return null; E temp = root.element(); root = removehelp(root, root.key()); nodecount--; return temp; } /** @return Record with key value k, null if none exist. @param k The key value to find. */ public E find(Key k) { return findhelp(root, k); } /** @return The number of records in the dictionary. */ public int size() { return nodecount; } private E findhelp(BSTNode rt, Key k) { if (rt == null) return null; if (rt.key().compareTo(k) > 0) return findhelp(rt.left(), k); else if (rt.key().compareTo(k) == 0) return rt.element(); else return findhelp(rt.right(), k); } /** @return The current subtree, modified to contain the new item */ private BSTNode inserthelp(BSTNode rt, Key k, E e) { if (rt == null) return new BSTNode(k, e); if (rt.key().compareTo(k) > 0) rt.setLeft(inserthelp(rt.left(), k, e)); else rt.setRight(inserthelp(rt.right(), k, e)); return rt; } /** Remove a node with key value k @return The tree with the node removed */ private BSTNode removehelp(BSTNode rt,Key k) { if (rt == null) return null; if (rt.key().compareTo(k) > 0) rt.setLeft(removehelp(rt.left(), k)); else if (rt.key().compareTo(k) < 0) rt.setRight(removehelp(rt.right(), k)); else { // Found it if (rt.left() == null) return rt.right(); else if (rt.right() == null) return rt.left(); else { // Two children BSTNode temp = getmin(rt.right()); rt.setElement(temp.element()); rt.setKey(temp.key()); rt.setRight(deletemin(rt.right(.

Objective Binary Search Tree traversal (2 points)Use traversal.pp.pdf

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On January 1, 2017, Pinnacle Corporation exchanged $3,527,500 cash for 100 percent of the outstanding voting stock of Strata Corporation. On the acquisition date, Strata had the following balance sheet: Cash $ 284,000 Accounts payable $ 383,000 Accounts receivable 311,000 Long-term debt 3,060,000 Inventory 443,000 Common stock 1,500,000 Buildings (net) 1,920,000 Retained earnings 1,205,000 Licensing agreements 3,190,000 $ 6,148,000 $ 6,148,000 Pinnacle prepared the following fair-value allocation: Fair value of Strata (consideration transferred) $ 3,527,500 Carrying amount acquired 2,705,000 Excess fair value $ 822,500 to buildings (undervalued) $ 480,000 to licensing agreements (overvalued) (110,000 ) 370,000 to goodwill (indefinite life) $ 452,500 At the acquisition date, Strata’s buildings had a 10-year remaining life and its licensing agreements were due to expire in 5 years. At December 31, 2018, Strata’s accounts payable included an $86,600 current liability owed to Pinnacle. Strata Corporation continues its separate legal existence as a wholly owned subsidiary of Pinnacle with independent accounting records. Pinnacle employs the initial value method in its internal accounting for its investment in Strata. The separate financial statements for the two companies for the year ending December 31, 2018, follow. Credit balances are indicated by parentheses. Pinnacle Strata Sales $ (7,683,000 ) $ (3,328,000 ) Cost of goods sold 4,940,000 1,835,000 Interest expense 338,000 207,000 Depreciation expense 630,000 448,000 Amortization expense 638,000 Dividend income (35,000 ) Net income $ (1,810,000 ) $ (200,000 ) Retained earnings 1/1/18 $ (5,160,000 ) $ (1,560,000 ) Net income (1,810,000 ) (200,000 ) Dividends declared 600,000 35,000 Retained Earnings 12/31/18 $ (6,370,000 ) $ (1,725,000 ) Cash $ 282,000 $ 547,000 Accounts receivable 1,255,000 305,000 Inventory 1,295,000 1,565,000 Investment in Strata 3,527,500 Buildings (net) 6,060,000 2,084,000 Licensing agreements 1,914,000 Goodwill 408,000 Total assets $ 12,827,500 $ 6,415,000 Accounts payable $ (337,500 ) $ (950,000 ) Long-term debt (3,120,000 ) (2,240,000 ) Common stock (3,000,000 ) (1,500,000 ) Retained earnings 12/31/18 (6,370,000 ) (1,725,000 ) Total Liabilities and OE $ (12,827,500 ) $ (6,415,000 ) a. Prepare a worksheet to consolidate the financial information for these two companies. PINNACLE COMPANY AND SUBSIDIARY STRATA Consolidation Worksheet For Year December 31, 2018 Consolidation Entries Accounts Pinnacle Strata Debit Credit Consolidated Totals Sales $(7,683,000) $(3,328,000) Cost of goods sold 4,940,000 1,835,000 Interest expense 338,000 207,000 Depreciation expense 630,000 448,000 Amortization expense 638,000 Dividend income (35,000) Net income $(1,810,000) $(200,000) $0 Retained earnings 1/1/18 (5,160,000) (1,560,000) Net income (1,810,000) (200,000) 0 Dividends declared 600,000 35,000 Retained earnings 12/31/18 $(6,370,000) $(1,725,000) $0 Cash $282,000 $547,000 Accounts rec.

On January 1, 2017, Pinnacle Corporation exchanged $3,527,500 cash f.pdf

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Most of the functions of a cell membrane are performed by proteins cholesterol None of this glycolysis Select all of the groups of membrane proteins All of the statements are correct Solution 13) It is a transmembrane protein 14) Transmembrane protein pass through both sides of the lipid bilayer and protude from both extracellular and cytoplasmic side of the membrane. It acts as a gateway to permit transport of proteins across membrane. 17) Transmembrane proteins are amphipathic as they are present in the bilayer and also protrude out of the membrane 12) Membrane proteins can be peripheral, intergral and lipid anchored..

Most of the functions of a cell membrane are performed by proteins ch.pdf

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Objectives 1. Understand the design, implementation, usage and limitation of a Linked List ADT 2. Gain experience implementing abstract data t types using already developed data structures Recognize and practice possible linked list applications. 3. Overview In this project, you should write a simple line editor. Keep the entire text on a linked list, one line in a separate node. Start the program with entering EDIT file, after which a prompt appears along with the line number. If the letter I is entered with a number n following it, then insert the text to be followed before line n. If I is not followed by a number, then insert the text before the current line. If D is entered with two numbers n and m, one n, or no number following it, then delete lines n through m, line n, or the current line. Do the same with the command L, which stands for listing lines. If A is entered, then append the text to the existing lines. Entry E signifies exit and saving the text in a file. Here is an Example: EDIT testfile 1> The first line 2> 3> And another line 4>I 3 3 The second line 4> One more line 5> L 1> The first line 3 The second line 4> One more line 5 And another line 5> D 2 4> L // This is now line 5, not 3; // line 5, since L was issued from line 5 //line 4, since one line was deleted; 1 The first line 2> The second line 3> One more line 4> And another line 4> E // this and the following lines // now have new numbers Solution.

Objectives 1. Understand the design, implementation, usage and limita.pdf

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New Nirvana Ltd is a company controlled by the members of the hard r.pdf

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Malaria being spread by mosquitoes is an example of which type of disase transmission? Vehicle-borne Zeonosis Fontice-borne Vector-borne The disease- evoking power of a pathogen is called which of the following? Invasiveness Variability Virulence Communicability The likelihood of a pathogen or agent being trans-from one infected person to another susceptive person is referred to as which of the following? Invasiveness Variability Virulence Communicability Metabolic diseases are best defined by which of the following? A collective name that refers to a group of many diseases with one common characteristic- uncontrolled growth of mutated cells A lower level of mental, physical, or moral state than is normal Diseases existing at birth and often before birth, or that develop during the first month of life Caused by the body reacting to an invasion of or injury by a foreign object or substance Any of the diseases or disorders that disrupt the process of converting food to energy on a cellular level; affects the ability of the cell to perform critical biochemical reactions that involve the processing or transport of proteins (amino acids), carbohydrates (sugars and starches), or lipids (fatty acids) As a general rule, a disease is included on a state\'s list of notifiable diseases under what conditions? Causes serious morbidity or death Has the potential to spread Can be controlled with appropriate intervention All of the above conditions warrant reporting the disease. The natural course of communicable disease involves a susceptible host; a point of exposure; a cube clinical disease phase; the clinical disease phase; and a phase of recovery, disability, or death. Which phase is related to the latency period? Susceptibility phase Subclinical phase Clinical disease phase Recovery, disability, or death phase Which term refers to quarantine-type activities that are conducted on an inpatient basis in a hospital or nursing home? Quarantine Isolation Good hygiene Immunization Chemoprophylaxis What herd immunity threshold did Jonas Salk suggest would make it infeasible for there to be a polio epidemic? 55% 65% 75% 85% Solution 13. b)- zoonosis 14. c)-virulence 15. d)- communicability 16. e) 17. d) - all of the above conditions 18. b) - subclinical phase 19. b) - isolation 20. d) - 85%.

Malaria being spread by mosquitoes is an example of which type of dis.pdf

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In oxidative phosphorylation and phosphorylation, the enzyme ATP syntheses couples the synthesis of ATP to which one of the following processes? A. The facilitated diffusion of protons B. The splitting of water with the release of O_2 C. The excilation of the Chlorophyll reaction center D. The reduction of NADP\' through NADP* reductase E. carboxylation of RuBP using the enzyme rubisco During which process(es) does the splitting of carbon dioxide to form oxygen gas and carbon A. photosynthesis B. both photosynthesis and respiration C. respiration D. neither photosynthesis nor respiration E. photorespiration. Which of the following describes a property of water necessary for life? A. Water changes temperature rapidly. B. Water is non-cohesive. C. Water is most dense as a solid. D. Water has constant high pH. E. Water can dissolve many chemicals. What is the lowest level of biological organization that can perform all of the activities required for life? A. organelle B. molecule C. cell D. tissue E. organism Solution 30. The proton gradient generated during electron transport drives the synthesis of ATP through ATP synthase. This process is similar in both oxidative phosphorylation and photophosphorylation. Excitation of the chlorophyll reaction center, the splitting of water with the release of O2 and, the reduction of NADP+ through NADP+ reductase occur only during the process of photophosphorylation. Carboxylation of RuBP using the enzyme rubisco takes place during Calvin cycle. Therefore, option A is correct. The enzyme ATP synthase couples the synthesis of ATP to the facilitated diffusion of protons during oxidative phosphorylation, and photophosphorylation. 31. Photosynthesis is the process in which light energy is used to reduce carbon dioxide to yield carbohydrates. During respiration, oxygen is consumed and carbon dioxide is released. Therefore, option A is correct. The splitting of carbon dioxide to form oxygen gas and yield carbohydrates occurs during the process of photosynthesis. 32. Water helps to regulate the temperature of the body. Water does not have high ph. It has a neutral ph. Water can dissolve many chemicals. It helps to transport essential molecules and chemicals throughout the body. Therefore, option E is correct. 33. Cell is the lowest level of biological organization that can perform all of the activities required for life. Cells that perform the same function are grouped together to form tissues, An aggregation of different tissues for a specialized function is called as an organ. Therefore, option C is correct. Cell is the lowest level of biological organization that can perform all of the activities required for life..

In oxidative phosphorylation and phosphorylation, the enzyme ATP synt.pdf

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List and define the significance of the following proteins/structures involved in DNA replication: DNA polymerase III, DNA polymerase I, Primase, Single stranded binding proteins (SSBPs), Helicase, Ligase, Primers, Topoisomerase, Anti-parallel ds DNA, Leading strand of DNA, Lagging Strand of DNA, replication bubble, and deoxyribonucleotide tri-phosphates (dNTPs). Describe the process of DNA replication using all of the terms that were listed (and defined) above. Make sure you include the difference in how the leading strand and lagging strand are synthesized. Also, include any differences (covered in class) between bacterial DNA replication and eukaryotic DNA replication. Lastly, include a hand drawn diagram with each of these structures and enzymes in different colors to illustrate how this process occurs. Solution DNA Replication: it is the biological process in which two identical replicas of DNA are produced. It involves many steps and components. Protein/ structures involved in DNA replication are as follows: DNA polymerase III: it was discovered by Thomas Kornberg and Malcolm Gefter. It is an enzyme which has its role to play in prokaryotic DNA replication. It has high processivity which is the number of nucleotides added per binding event. It can also perform proof reading which is the correction of wrongly added nucleotides is DNA polymerase I: it was first discovered polymerase enzyme and Arthur Korenberg discovered it. It does not have a role in replication but poses exonuclease activity that is removal of nucleotides during DNA repair. Primase: it is a type of RNA polymerase. This is very important in initiation of DNA replication because no polymerase can synthesise a DNA without small single stranded RNA. Thus a primase help in synthesising such small single stranded RNA complementary to the template DNA. After initiation, this RNA piece is removed. Single-stranded binding proteins (SSBPs): these are the class of proteins that prevent premature binding of nucleotides to template DNA. They protect single stranded DNA from being digested by nucleases and also prevents looping of single stranded DNA, so that other enzymes can perform their function. Helicases: these are the group of enzymes which help in separating two strands of DNA for the process of replication. Ligases: another class of enzymes which help in binding of two strands of DNA by catalysing formation of phosphodiester bond between the two strands. Primer: a short single stranded DNA/RNA that serves as a starting point for the DNA synthesis. Primase helps in the formation of primer. Topoisomerases: during replication DNA over wounds ahead the replication point making further replication difficult, in such instance topoisomerase help in unwinding supercoiled DNA. Anti-parallel double strand: DNA is made of two strands running in opposite directions and hence anti parallel one runs from 5’ to 3’ and the other from 3’ to 5’. Leading strand: DNA replication on leading stra.

List and define the significance of the following proteinsstructures.pdf

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In this TEM image, what does the TEM allow you to view that would not be visible using the Compound Microscope? The human cell SELECT ALL ANSWERS THAT APPLYThe structure of internal cell organellesThe Human Immunodeficiency Virus The human cell SELECT ALL ANSWERS THAT APPLY Solution Answers are The structure of internal cell organelles And The human immuno deficiency virus. Transmission electron microscope TEM can be used to see extremely small things like cell organelles and viruses. With compound Microscope we cannot see those. Human cell can be seen with compound Microscope..

In this TEM image, what does the TEM allow you to view that would no.pdf

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