Question 6 (10 points) The following information is from reformulated.pdf

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Question 6 (10 points) The following information is from reformulated financial statements (in millions of dollars): 2007 2008 Net operating assets (NOA)1,9002,400 Net financial obligations (NFO)1.000 A. Calculate ROCE for 2008; use average book value in the denominator. Solution ROCE = Net Income / Average Capital employed 2007 2008 Average Net Operating Assets 1900 2400 2150 Net Financial Obligations 1000 1200 1100 NOA-NFO (Capital Employed) 900 1200 1050 Since you have not provided the complete data, I’ve referred to the same question in our database and used the information. ROCE = 207/1,050 = 0.1971 or 19.71% 2007 2008 Average Net Operating Assets 1900 2400 2150 Net Financial Obligations 1000 1200 1100 NOA-NFO (Capital Employed) 900 1200 1050.

Question 6 (10 points) The following information is from reformulated financial statements (in
millions of dollars): 2007 2008 Net operating assets (NOA)1,9002,400 Net financial obligations
(NFO)1.000 A. Calculate ROCE for 2008; use average book value in the denominator.
Solution
ROCE = Net Income / Average Capital employed
2007
2008
Average
Net Operating Assets
1900
2400
2150
Net Financial Obligations
1000
1200
1100
NOA-NFO (Capital Employed)
900
1200
1050
Since you have not provided the complete data, I’ve referred to the same question in our
database and used the information.
ROCE = 207/1,050 = 0.1971 or 19.71%
2007
2008
Average
Net Operating Assets
1900
2400
2150
Net Financial Obligations
1000
1200
1100

NOA-NFO (Capital Employed)
900
1200
1050

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describe these organelles and the organisms there found in 1. crystalline rod 2. kinetoplast 3. apicocomplex 4. contractile vacuole Solution 1. Crystalline rod: Some protozoansort called euglenozoans contain spiral or crystalline rod inside their flagella. Two examples of euglenozoans with these rods arekinetoplastidsandeuglenids. The flagella of kinetoplastids and euglenids have paraxial rods running next to each other. 2.Kinetoplast: A kinetoplast is a network of circular DNA (called kDNA) inside a large mitochondrion that contains many copies of the mitochondrial genome.A kinetoplast is usually adjacent to the organism\'s flagellar basal body, suggesting that it is tightly bound to the cytoskeleton. Kinetoplasts are only found in protozoa of the class Kinetoplastida. 3.Apicocomplex: It is an organ complex of the Apicomplexa that appears as a conical structures on the tapered end (or the apical end) of the cell, and contains rhoptries, micronemes, polar rings, and conoid. The organelle is an adaptation that the apicomplexan applies in penetration of a host cell. The organelle is found in a large phylum of parasitic alveolates called Apicomplexa. The Apical complex consists of a set of spirally arranged microtubules (the conoid), a secretory body (the rhoptry) and one or more polar rings. Additional slender electron-dense secretory bodies (micronemes) surrounded by one or two polar rings may also be present. This structure gives the phylum its name. 4.contractile vacuole: A contractile vacuole (CV) is a sub-cellular structure (organelle) involved in osmoregulation. It is found predominantly in protists and in unicellular algae. The contractile vacuole acts to regulate the quantity of water inside of a cell. In freshwater environments, the concentration of solutes inside the cell is of a higher concentration than outside the cell (i.e., the environment is hypotonic). Under these conditions, water flows from the environment into the cell by osmosis. The contractile vacuole acts as part of a protective mechanism that prevents the cell from absorbing too much water and possibly lysing (rupturing) through excessive internal pressure..

describe these organelles and the organisms there found in1. cryst.pdf

describe these organelles and the organisms there found in1. cryst.pdf

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Compare the small intestine to the large intestine in terms of histological (tissue) organization, size, and unique anatomical features. How do the differences between the small and large intestines give you a clue as to their different functions in the digestive system? Solution Small intestine: it is the largest part of the alimentary canal with narrow diameter and it\'s anatomical features are it contains the three parts namely duodenum, jejunum and ileum.Jejunum is the middle portion between the duodenum and ileum.Duodenum contains the pancreas and small intestine receives the pancreatic juices and bile .It contains the peyers patch in small intestine and haustoria are absent.Internal surfaces contains the circular folds or plicae circulates. villa are present. Brunner\'s glands are present. size ranges from 4.5 to 7 m long and width of 3.5 to 4.5 width. Histological features are it contains four layers like serosa, muscularis,sub mucosa, mucosa.Serosa is the outer most layer and it secrets serous fluid, muscularis is responsible for gut movement that is peristalsis, sub mucosa joins to the mucosa and underlying smooth muscle. The main function of small intestine is to absorb nutrients and minerals. Large intestine: It is comparatively shorter than small intestine .It\'s size ranges from 1.5 meters long and 4 to 5 cm in diameter. It\'s anatomical features are it has ascending colon consists of caecum and appendix, transverse colon consists of colic flexures and transverse mesocolon, descending colon and sigmoid s shaped colon. It\'s histological features are thick mucosa with deep crypts but it does not contain villi and the epithelium consists of columnar absorptive cells, endocrine cells basal stem cells and many goblet cells. The main function of large intestine is to absorb water from remaining indigestible food and transmit the useless waste materials from the body. As they have different anatomical , histological and structural features so we can say that the small intestine and large intestine are different in their respective functional features..

Compare the small intestine to the large intestine in terms of histo.pdf

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Chemical composition of rain water. Solution Rainwater gets its compositions largely by dissolving particulate materials in the atmosphere (upper troposhere) when droplets of water nucleate on atmospheric particulates, and secondarily by dissolving gasses from the atmosphere. Rainwater compositions vary geographically. In open ocean and coastal areas they have a salt content essentially like that of sea water (same ionic proportions but much more dilute) plus CO2 as bicarbonate anion (acidic pH). Terrestrial rain compositions vary siginificantly from place to place because the regional geology can greatly affect the types of particulates that get added to the atmosphere. Likewise, sources of gaesous acids (SO3, NO2) and bases (NH3) vary as a function of biome factors and anthopogenic land use practices. Each of these gasses can be added in varying proportions from natural and non natural input sources (non-natural sources of SO3 and NO2 far outweigh natural ones).

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An adult male receives all his water from a municipal supply that is contaminated with 0.003mg/Lof tetrachloroethylene (PCE). Assume he takes a 15-minute shower every day. The PCE concentration in the air is in equilibrium with the PCE in the water and thus can be determined by Henry’s law from the concentration in the water. PCE has a unitless Henry’s constant KH=1.4, which means Cair=Caq/KH. The dermal permeability constant of PCE is 6.17×10-5 m/min. (10 pts) a. Calculate his total annual (ED=AT=1 yr) LADD due to exposures of drinking the water and showering (dermal absorption + inhalation) with the water. b. Which of the three modes of exposure results in the highest delivered dose? Solution There are generally three routes of exposure: dermal, oral and inhalation, as clear from the question. Intake from these routes of exposure differ based upon the chemical and physical properties of the toxicant as well as the permeability of the surface at the site of absorbtion, here, lung, skin and the GI tract. LADD (Lifetime Average Daily Dose) is obtained by adding up the AADD(Average Adsorbed Daily Dose) for all the years of a person\'s life, namely: ->infant (1yr), ->child (1-6yrs), ->child (7-12 yrs), ->adolescents (13-18yrs), ->adults (19-70yrs) AADD is obtained by adding adsorbed dose (AD) through all the 3 ways. AD=Exposure x Adsorption Factor Exposure = Concentration x Intake x Duration x Frequency Body Weight _____________________________________________________________________________ ____ Another way to calculate LADD is: LADD = (C x IR x ED)/(BW x LT) ; where; C = concentration of chemical IR = intake rate ED = exposure duration BW = body weight LT = lifetime (also AT) (a) Total Annual LADD due to drinking water + showering (1) Drinking water C = 0.003 mg/L IR = 2 L daily; for 1 year = 2*365 = 730 L ED = AT = 1 year = 365 days LT = 70 years = 25,550 days BW = 70 kg i.e, LADD = (0.003 * 730 * 1) / (70 *365) = 8.57 x 10-5 mg/kg.day (2) Similarly, for dermal absorption For dermal adsorption, Dosage, D = (C x P x SA x ET x CF) / (BW); where D = Dose ( mg/kg day) C = conc. of contaminant (mg/L) P=Permeability coefficient (cm/hr) SA = exposed body surface area (cm2); here 19400 cm2 (Standard value) ET = Exposure time (hrs/day); here 0.25 hours /day CF = conversion factor = 1 L/1000 cm3 BW = body weight = 70 kg thus, D= (0.003*(6.15*10-5*10-2/60)*19400*0.25))/(1000*70) =2.13*10-12 mg/kg.day. However for 1 year, D = 7.776 * 10-10 mg/kg.day Thus, total LADD = 8.57*10-5 + 7.776 * 10-10 = 8.57 * 10-5 mg/kg.day (b) Usually, studies show that of all the modes of exposure, ingestion or drinking the water will deliver the highest dose. Here, dosage due to air inhalation will be 8.57 * 10-5 / 1.4 = 6.121 * 10-5 mg/kg.day (as Cair = Caq/1.4) Thus, comparing all the values, drinking the water will cause more dosage..

An adult male receives all his water from a municipal supply that is.pdf

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C++ problem Part 1: Recursive Print (40 pts) Please write the recursive List reverse print function, whose iterative version we wrote in class. Below are the function signatures for the functions you are going to need: public: /**Additional Operations*/ void print_reverse(); //Wrapper function that calls the reverse helper function to print a list in reverse //prints nothing if the List is empty private: void reverse(Nodeptr node); //Helper function for the public printReverse() function. //Recursively prints the data in a List in reverse. Why do we need the private helper function here? Since we are going to be reversing our list node by node, in a recursive fashion, we want to pass a one node at a time to our reverse function. However, since our nodes are private, we cannot access them if we call the function inside of main. Add these function signatures to your List.h file along with your other function prototypes inside the class definition. Make sure that you place the reverse function inside the private portion of your List class definition and the print_reverse function prototype to the public portion of your List class definition. Now, implement these two functions inside of List.h, under your section for additional operations. Important: Test each function carefully inside of your ListTest.cpp to make sure that it is working properly. Part 2: Adding an Index to Your List Nodes (20 pts) Next, you will add the following functions to your List.h /**Accessor Functions*/ int get_index(); //Indicates the index of the Node where the iterator is currently pointing //Nodes are numbered from 1 to length of the list //Pre: length != 0 //Pre: !off_end() ... int List::get_index() { //Implement the function here } /**Manipulation Procedures*/ void scroll_to_index(int index); //Moves the iterator to the node whose index is specified by the user //Pre: length != 0 ... void scroll_to_index(int index) { //Implement function here } Part 3: Implementing Search as Part of Your List (40 pts) Now, we are going to add two search functions to our List so that we can search for elements in our List. The first of these functions is going to be a simple linear search function. You will need to add the following function prototype and function definition to your List.h: /**Additional Operations*/ int linear_search(listitem item); //Searchs the list, element by element, from the start of the List to the end of the List //Returns the index of the element, if it is found in the List //Returns -1 if the element is not in the List //Pre: length != 0 ... int List::linear_search(listitem item) { //Implement the function here } You are also going to add a function to perform recursive binary search on your List. You will need to add the following function prototype and function definition to your List.h: int binary_search(int low, int high, listitem item); //Recursively searchs the list by dividing the search space in half //Returns the index of the element, if it is fo.

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1.What does the emergence of autonomous vehicles mean to you? What value about car travel would you increase, decrease, eliminate? 2. Think of at least 3 Blue Ocean innovations that would be needed in order for autonomous vehicles to be successful: these can be internal/organizational efficiency innovations or new products/services. 3. Do you think the biggest barriers will be technical or social in nature? 4) each individual should create a list of at least 10 “factors of competition” that are relevant to the personal transportation market. 5) Conduct “mini-interviews” with at least 5 people and get feedback about your identified factors of competition – are some factors more important to people than others, are there factors identified by your interviewees that you missed? Note: these are informal “mini-interviews”, and each individual must conduct at least 5 interviews to provide market feedback on your factors of competition. Record your notes from each mini-interview using the table structure below: Interviewee Name, Age, Location, Gender and About Eliminate Reduce Raise Create Insights Mary Jones: 48 from Birmingham, AL. Female. Busy working mom with 2 middle-school kids 1. External Appearance 2. Taxi services 3. New road construction 4. Road signs 5. Mom having to drive kids to soccer practice – YEAH!! 1. Complexity of the dashboard 2. Air pollution 3. Engine size and performance 4. Car salesmen and personal car ownership 1. Entertainment and internet access in car 2. Comfort through alternative seating styles (e.g. 4 people facing each other around a table) 3. Ease of maintenance and longevity of cars 1. Members-only service (e.g. I belong to the Volvo driving club, which has its own app and fleet of cars) 2. Vehicle support services – cars need to come in to get gas and fixed by mechanics Value proposition: Time (not having to drive kids everywhere) Safety (people could drink and get home safely) Fun (spend travel time playing with kids, watching movies etc.) individual’s interview summary table, each individual’s original list of 10 competitive factors. These should all be aggregated into one document and uploaded to our Canvas course site. Interviewee Name, Age, Location, Gender and About Eliminate Reduce Raise Create Insights Mary Jones: 48 from Birmingham, AL. Female. Busy working mom with 2 middle-school kids 1. External Appearance 2. Taxi services 3. New road construction 4. Road signs 5. Mom having to drive kids to soccer practice – YEAH!! 1. Complexity of the dashboard 2. Air pollution 3. Engine size and performance 4. Car salesmen and personal car ownership 1. Entertainment and internet access in car 2. Comfort through alternative seating styles (e.g. 4 people facing each other around a table) 3. Ease of maintenance and longevity of cars 1. Members-only service (e.g. I belong to the Volvo driving club, which has its own app and fleet of cars) 2. Vehicle support services – cars need to come in to get gas and fixed by mechanic.

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You have installed a PCIe wireless NIC in a computer desktop in a small office environment that cannot be wired for an Ethernet network. Despite the fact that the nearest access point is well within range, the computer still receives a poor Wi-Fi signal. What can you do to improve the computer’s reception? A. Turn up the sensitivity on the PCIe wireless NIC. B. Install a PCIe wireless NIC with a stronger receiver. C. Install an external USB wireless NIC that can have its antenna placed to get better reception. D. Change the entire wireless network to use an 802.11 standard that has greater range. Solution C. Internal wireless NICs have the antenna directly connected to the NIC so that it sits at the back of the computer. This might not be the ideal location to receive Wi-Fi signals, even if the computer is within range otherwise. Using a USB wireless NIC with an external antenna that can be moved for better reception is a more workable option..

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You have identified a novel enzyme. Supercoiled DNA shows an upward mobility shift following a short incubation with the enzyme. Following a long incubation with the enzyme, the supercoiled DNA seems to disappear or be converted in a large number of small fragments. What kind of enzyme is our novel enzyme likely to be? Type I topoisomerase Type II topoisomerase Restriction enzyme DNA nuclease A and B are both correct Solution .C. Restriction enzymes . are a class of enzymes that act upon the DNA molecule at specific site and cleave DNA molecule into small fragments..

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Which of the following wireless standards uses direct sequence spread spectrum (DSSS) by default? A. Bluetooth B. 802.11a C. 802.11b D. 802.11ac Solution C. 802.11b uses direct sequence spread spectrum (DSSS) technology. DSSS is a transmission method that transmits the data along with a chipping bit to increase the signal’s resistance to interference. Answer A is incorrect because Bluetooth uses frequency-hopping spread spectrum. Answer B is incorrect because 802.11a uses orthogonal frequency-division multiplexing. Answer D is incorrect because 802.11ac uses MIMO-OFDM.

Which of the following wireless standards uses direct sequence sprea.pdf

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Which one of the following statements about lichens is false? lichens are resistant to pollution. Lichens can withstand severe drought. lichens are important pioneers on new land. Some lichens are thousands of years old. Some lichens tolerate severe cold. Which of the following bacterium type would die in the presence of oxygen(poisoning)? Obligate aerobes Obligate anaerobes Facultative anaerobes All of the above Which of the following traits do archaeans and bacteria share? composition of the cell wall. presence of plasma membrane lack of a nuclear envelope identical rRNA sequences 1 only 3 only 1 and 3 2 and 3 2 and 4 What do \"fimbriae\" of bacteria do? Catch food Absorbing liquid food Sticking to a substrate. Help in sexual reproduction All of the above. Picking up naked exracellular DNA molecule is accomplished during Transcription Conjugation Transformation Transduction Translation Which of the following confers antibiotic resistance? Transduction Jumping genes Hfr R plasmid F plasmids Solution Lichens are resistant to pollution Obligate anaerobes lack of nuclear envelope Sticking to the substrate transformation R plasmid.

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Which one of the following statements about lichens is false lichen.pdf

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What is the capability to combine data from separate sources to gain information? A. Metadata B. Inference C. Aggregation D. Deadlocking Solution C. Aggregation is the capability to combine data from separate sources to gain information. Answer A is incorrect because metadata is data about data. Answer B is incorrect because inference attacks occur when authorized users infer information by analyzing the data they have access to. Answer D is incorrect because deadlocking is a database stalemate.

What is the capability to combine data from separate sources to gain.pdf

What is the capability to combine data from separate sources to gain.pdf

What is the capability to combine data from separate sources to gain.pdf

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What are Factors that influence primary production in different zones of oceans and lakes Solution Organisms that synthesise the organic elements of life from the inorganic molecules are known as \"primary producers.\" Plants, certain bacteria and algae are autotrophs (they feed themselves by photosynthesis). Photosynthesis is a process through which plants use sun energy for the production of organic material. Plants utilise necessary amount of sunlight for photosynthesis. The major factors affecting the primary production in different zones of oceans and lakes are, 1). Availability of Sun light 2). Mineral nutrients In contrast, the terrestrial plants have the limitations of water supply and uniform temperature. As the aquatic plants and organisms live in the water itself, they do not suffer from water scarcity. Also, temperature is uniform due to the heat capacity of water..

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Using NetBeans Q1. Write a program that uses a stack to convert a number in base 10 to a number in base 2. Solution import java.util.Scanner; public class Decimal2BinaryDemo { public static void main(String args[]) { Scanner in =new Scanner(System.in); System.out.println(\"Enter decimal number =\"); int num=in.nextInt(); Stack s4 = new Stack(); int n=num; int i = 0; while (n > 0) { s4.push(n % 2); n = n / 2; i++; } //s4.print(); System.out.print(num+\" is convert to binary ->\"); while (i > 0) { System.out.print(s4.pop()+\" \"); i--; } } }.

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Use the counting techniques from the last chapter. A bag contains four red marbles, two green ones, one fluorescent pink one, three yellow ones, and three orange ones. Suzan grabs five at random. Find the probability of the indicated event. She gets all the red ones, given that she gets the fluorescent pink one. Solution Given that she gets the fluorescent pink one, she needs to pick 4 more from remaining 12 marbles Out of these 4 more, she can pick 4 red marbles in 4C4 ways. Total number of ways of selecting 4 marbles from 12: 12C4 Probability of all red picked up = 4C4 / 12C4 = 1/495 = 0.00202.

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Use the counting techniques from the last chapter. A bag contains fo.pdf

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the three types of associations (chance, non-causal, and causal). Describe each giving an example of each . Solution Association refers to linkage among variables. 1.)Chance:It is a fake or false association due to bias.For example:increased coke intake and crime rate in a country. 2.)Casual:Cause and effect relation can be seen that is independent variable must cause change in independent variable. For example: sugar intake and high diabetes. 3.)No direct association between two variables.For example: alcohol use and smoking..

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The problem of so-called “push polls” When advocacy calls are made under the guise of research This statement from AAPOR, American Association for Public Opinion Research, explains how to tell the difference between fraudulent political polls—commonly referred to as “push polls”—and legitimate polling, including message testing. AAPOR condemns political telemarketing under the guise of research and is committed to providing information that explains what this unethical campaign practice is and what you can do about it. A “Push Poll” is not a Legitimate Poll A so-called \"push poll\" is an insidious form of negative campaigning, disguised as a political poll. “Push polls” are not surveys at all, but rather unethical political telemarketing -- telephone calls disguised as research that aim to persuade large numbers of voters and affect election outcomes, rather than measure opinions. This misuse of the survey method exploits the trust people have in research organizations and violates the AAPOR Code of Professional Ethics and Practices. Identifying Advocacy Calls Made Under the Guise of Research Political telemarketing calls, when disguised as research, may sometimes be difficult to differentiate from a legitimate survey. Here are characteristics that will usually indicate to a respondent that the call is not a legitimate survey. One or only a few questions are asked, all about a single candidate or a single issue. The questions are uniformly strongly negative (or sometimes uniformly positive) descriptions of the candidate or issue. The organization conducting the calls is not named, or a phony name is used. Evasive answers are given in response to requests for more information about the survey. In addition, the following characteristics will indicate to journalists, reporters, and survey professionals that a telephone call is not a legitimate survey. The number of people called is very large, sometimes many thousands. The calls are not based on a random sample. It is difficult to find out which organization conducted the interviews. Fraudulent Polls vs. Message Testing The fact that a poll contains negative information about one or more candidates does NOT in and of itself make it a \'push poll.’ Political campaigns routinely sponsor legitimate “message- testing” surveys that are used by campaign consultaest out the effectiveness of various possible campaints to tgn messages or campaign ad content, often including negative messages. Political message-testing surveys may sometimes be confused with fake polling, but they are very different. One way to tell is that message-testing surveys exhibit the characteristics of a legitimate survey, such as: At the beginning of the call, the interviewer clearly identifies the call center actually making the calls. (However, legitimate political polling firms will often choose not to identify the client who is sponsoring the research, be it a candidate or a political party, since that could bias the survey .

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The phylogeny in this figure shows the evolution of visual opsins in tetrapods. Each colored triangle represents a different opsin. Which of the following accurately describes the opsins in these species Squamate reptiles Birds New World Humans primates A opsin gain A Opsin loss opsin loss Tetrapod ancestor Solution Synapomorphy is a character present in ancestral species and shared exclusively by its evolutionary descendants Symplesiomorphy is an ancestral character that is shared by two or more taxa. Based on the choices given above the last choice is the correct answer. The green opsin was developed in the ancestors of the baboons and human and hence the two groups share that character. The light blue opsin was present in the tetrapod ancestor and it is now shared by the two taxa descended from the common ancestor..

The phylogeny in this figure shows the evolution of visual opsins in .pdf

The phylogeny in this figure shows the evolution of visual opsins in .pdf

The phylogeny in this figure shows the evolution of visual opsins in .pdf

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The hypothetical Western Chickadee comprises three major populations: Pacific, Great Basin, and Rockies. In the chickadees, the bill color gene has two alleles (B_1 & B_2). Relevant information about genotypes and the populations follows: Pacific N =- 350,000 f (B_1) = 0.97 Great Basin N = 450,000 f (B_1) = 0.96 Rockies N = 750,000 f (B_1) = 0.98 B_1 B_1 bill color = black, relative fitness = 1.0 B_1 B_2 bill color = black, relative fitness = 1.0 B_2 B_2 bill color = yellow, relative fitness = 0.6 In which population is drift likely to have the largest effect? Pacific Great Basin Rockies There is not enough information provided to determine this. Solution Hi, I hope my answer is satisfactory. Thanks!!.

The hypothetical Western Chickadee comprises three major populations.pdf

The hypothetical Western Chickadee comprises three major populations.pdf

The hypothetical Western Chickadee comprises three major populations.pdf

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The accumulation of partially decayed plant material is called Pea Peat may be covered with sediment and placed under extreme andss With time and pressure peat may be compressed into boaAu tread (Paragraph 2)The rate of fossil fuel formation was great during the period. The lands were covered by Carbon accross and the climate was ool Zdanne so this favored year round growth. The Svps dominated the landscape and hence created most of the coal. Also present but not in type of seed plants. abundance were the (Paragrpah 3) which tree dominated in the Carboniferous? trees Another tree, Giant Horsetails, was also abundant. These were both svps, so they still had external fertilization (released spores and sperm into the water). They flourished in this tropical/wet climate. (Pg 401, paragraph 2) What happened to the climate and hence the svp trees at the end of the \"R Carboniferous? climakt tin coz t ceel ca med agey in The water receded so the spores and sperm croaked Only pmaALlycDonytes relatives of SVP trees remain today. One group of Gymnosperms, survived to today. They evolved pollen to transport sperm and so moved to internal fertilization. They were able to take over the drying landscape at the end of the Carboniferous!!!!! Solution Ques-1: The accumulation of partially decayed plant material is called peat. Peat may be covered with sediment and placed under extreme pressure. With time and pressure peat may be compressed into \"peat bog\" or mire Ques-2 The rate of fossil fuel formation was great during the carboniferous period. The lands were covered by peat (dead trees & dead plant organisms) and climate was cold & damped. So this favored year round growth. The SVPs dominated the landscape and hence created most of the coal. Also present but not in abundance were the vascular -angiosperm type of seed plants.

The accumulation of partially decayed plant material is called Pea Pe.pdf

The accumulation of partially decayed plant material is called Pea Pe.pdf

The accumulation of partially decayed plant material is called Pea Pe.pdf

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19. DNA “Nucleotide bases” are selective in their bonding. Thinking about your readings, and the details discussed in lecture 2~4, what is the corresponding (“mirror image”) sequence of bases for the DNA strand CTAATGT? In question 19, I stated that the sequence of nucleotide bases was DNA, rather than RNA, but you could have known this anyway, just by looking at the bases that are present. How do you know that this was not a sequence of RNA? Solution Genes consist of specific sequences of DNA nucleotides (complementary base pairs) which code for the structure of polypeptide chains. These polypeptide chains are tRNA (transfer RNA) and mRNA (messenger RNA), which perform transcription and translation processes respectively during protein synthesis. So, these DNA nucleotides otherwise called “bases” carry the message encoding for the synthesis of a protein. There are four such bases, namely, adenine (A), thymine (T), guanine (G), and cytosine (C), which are the building blocks of DNA’s double helical structure. RNA contains Uracil (U) in place of thymine. Among these base pairs, adenine binds with thymine and guanine binds with cytosine, this is called as complementary pairing. In RNA, adenine binds with uracil. The corresponding sequence/mirror image/complimentary sequence od the given DNA sequence CTAATGT is, GATTACA. We can determine that the above sequence is DNA because thymine is present. RNA contains uracil in place of thymine..

19. DNA “Nucleotide bases” are selective in their bonding. Thinking .pdf

19. DNA “Nucleotide bases” are selective in their bonding. Thinking .pdf

19. DNA “Nucleotide bases” are selective in their bonding. Thinking .pdf

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describe these organelles and the organisms there found in1. cryst.pdf

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