Derived characteristics that define the order Primate include a.pdf

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Derived characteristics that define the order Primate include a. highly developed olfaction (sense of smell). b. increased dependence on complex behavior, learning, and behavioral flexibility. c. specialized molars, up to six incisors, and a lack of premolars. d. bipedalism. Solution In animal order primate including lemurs,tarsiers,monkeys,apes and human beings. So all these animals have complex beahvior,learning and beahvioral felxibility which is common feature in all.Primates have limb structure, one upper limb bone, two lower limb bones, many mammalian orders have lost various bones, especially fusing of the two lower limb bones.primates have the internal regulation of body temperature and feeding their young with milk produced by mammary glands..

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The portion of the fungus that projects above the substrate and contain the spores is known as A) septate hyphae B) septate mycellum C) aerial mycellum D) vegetative mycelium E) none of the above = identity Bacteria contamination is suppressed in plain Sabouard\'s agar by the addition of peptone and dextrose B) brain heart infusion media C) Lactophenol Cotton Blue D) all of the above E) none of the above = identity Solution 16. C. Aerial mycelium - the portion of the mycelium that project above the substrate and produce spores. Because of its capability to produce spores, it is also known as reproductive mycelium. 17. D. all of the above.

The portion of the fungus that projects above the substrate and conta.pdf

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the appearance of symptoms of a genetic disorder at earlier ages in .pdf

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Stuck on hw problem Assume that the forces acting on a mass are such that the potential energy is the function of x shown in Fig. 24-5: Sketch the solutions in the phase plane. Describe the different kinds of motion that can occur. Solution Every sound we hear, every photon of light that hits our eyes, the movement of grass blown by the wind and the regular beat of the tides are all examples of waves. They are all around us. Visible, physical wave*s such as those we see when a rock is thrown into water are what many people think about when they first began to think about waves. These waves have distinct properties specific to their type but also exhibit characteristics in common with more abstract waves such as sound waves and light (electromagnetic) waves. Below is an interactive illustration that explores particle and energy motion for the three wave types listed above. Once you have reviewed the illustration and accompanying text, test your understanding of these concepts with the wave type At the simplest level, waves are a disturbances that propagate energy through a medium. Propagation of the energy depend on interactions between the particles that make up the medium. Particles move as the waves pass through but there is no net motion of particles. This means, once a wave has passed the particles return to their original position. As a result, energy, not matter, is propagated by waves. Different types of waves exhibits specific characteristics. These characteristics are used to distinguish between wave types. Orientation of particle motion relative to the direction of energy propagation is one way waves are characterized. There are three categories: A second way that waves are characterized is by the types of matter they are able to travel through.Electromagnetic waves (including light) can move through a vacuum. Physical waves require matter through which to propagate. Physical waves are further distinguished by the phases of matter through which they can move. Longitudinal waves can pass through liquids and gasses while transverse waves require a material to be solid to propagate..

Stuck on hw problem Assume that the forces acting on a mass are such.pdf

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Some Giant Sulphur Bacteria are so large as to be visible to the naked eye. Nonetheless, they have been very difficult to cultivate and classify. Why? Solution Cultivate Grow very slowly. Metabolism is based on lithotrophic utilization of sulphide oxidation coupled to nitrate reduction which is gives poor yields. In natural environments they grow in gradients where one end of filament is sulphide rich, the other end nitrate rich. Difficult to simulate in the lab. Classify Morphology has been used, but this is not linked to phylogeny. 16S rRNA hard to amplify due to containing an intron.

Some Giant Sulphur Bacteria are so large as to be visible to the nak.pdf

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right now I am using a l293d chip to control my DC motor which is 6 volt. I was wondering if there was a python code that can control the motor based on temperature. I am using tmp102 ic2 temperature sensor also. Solution L293D IC generally comes as a standard 16-pin DIP (dual-in line package). This motor driver IC can simultaneously control two small motors in either direction; forward and reverse with just 4 microcontroller pins (if you do not use enable pins). Some of the features (and drawbacks) of this IC are:.

right now I am using a l293d chip to control my DC motor which is 6 .pdf

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Problem 1: Eclipses and occults Saturn has five satellites that rotate around the planet. The list below records events occurring during the month of August, 1995, involving eclipses or occults of Saturnian recorded satellites. For each event, the percent of light lost by the eclipsed or occulted satelite is Event Aug 2 Aug 4 Aug 5 Aug 6 Aug 8 Light Loss (%) 61 56 Eclipse Eclipse Occult Eclipse Eclipse Occult Occult Occult Occult Occult Eclipse Occult Occult Occult Occult Occult Occult Occult Occult 46 Aug 11 39 Aug 12 Aug 14 Aug 14 Aug 15 Aug 15 AUg 16 Aug 20 Aug 23 Aug 23 AUg 25 Aug 28 1 20 12 Source: Astronomy magazine. Aug. 1995, p.60 via Statistics (10m ed.] McClave and Sincich, p. 53 Solution a). Stem and Leaf plot for light loss stem unit = 10 leaf unit = 1 Frequency Stem Leaf 8 0 1 1 2 3 4 5 5 9 3 1 1 2 3 2 2 0 0 1 3 9 1 4 6 1 5 6 2 6 1 5 0 7 0 8 0 9 1 10 0 19 b). Stem and Leaf plot for light loss stem unit = 10 leaf unit = 1 Frequency Stem Leaf 8 0 1 1 2 3 4 5 5 9 3 1 1 2 3 2 2 0 0 1 3 9 1 4 6 1 5 6 2 6 1 5 0 7 0 8 0 9 1 10 0 19 (eclipse are marked red) c). Based on above graph, eclipse is more likely lead to greater light loss. Stem and Leaf plot for light loss stem unit = 10 leaf unit = 1 Frequency Stem Leaf 8 0 1 1 2 3 4 5 5 9 3 1 1 2 3 2 2 0 0 1 3 9 1 4 6 1 5 6 2 6 1 5 0 7 0 8 0 9 1 10 0 19.

Problem 1 Eclipses and occults Saturn has five satellites that rotat.pdf

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please show work A hospital reports that two patients have been admitted who have contracted Crohn\'s disease. Suppose our experiment consists of observing whether each patient survives or dies as a result of the disease. The simple events and probabilities of their occurrences arc shown in the table (where 5 in the first position means that patient 1 survives, D in the first position means that patient 1 dies, etc.). Solution.

please show work A hospital reports that two patients have been admi.pdf

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Person.java is given Solution Since I didn\'t have the implementation of the Disease class, I made some assumptions and wrote the code accordingly. public class Person{ public final String name; protected int health; protected Disease[] diseases; public Person(String name, int health){ this.name = name; this.health = health; } public int getHealth(){ return this.health; } public void addDisease(Disease dis){ if (!diseasePresent(dis)){ Disease[] updated_diseases[diseases.length+1] = new Disease[]; for (int i<0; i.

Person.java is givenSolutionSince I didnt have the implement.pdf

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In java language Which of the following is false? A) A JPasswordField shows that characters are being entered, but hides the actual character with an echo character. B) Class JTextField extends JTextComponent; Class JPasswordField extends JTextField. C) Both JTextFields and JPasswordFields are single-line areas in which the user can enter text via the keyboard. D) JPasswordFields generate events; JTextFields do not. When the user presses Enter in a JPasswordField, the GUI component generates an ________, which is processed by an object that implements the interface _______. A) ActionEvent, ActionListener B) ActionEvent, ActionEventListener C) TextEvent, TextListener D) TextEvent, TextEventListener JPasswordField contains method ___________to obtain the data entered. A) getInput B) getPassword C) getText D) getEcho A JRadioButton is different from a JCheckBox in that _________. A) a JRadioButton is a subclass of JToggleButton, JCheckBox is not B) normally several JRadioButtons are grouped together and are mutually exclusive C) a JRadioButton is a type of button, JCheckBox is not D) a JRadioButton is a state button, JCheckBox is not The fact that class Graphics is abstract contributes to Java’s portability because: A) drawing is performed differently on every platform that supports Java. A subclass of Graphics must be created that uses the drawing capabilities of the current platform. B) objects of non-abstract classes can only be instantiated on the Windows platform. C) drawing should not be performed on non-Linux platforms. D) Class Graphics is not abstract. JSliders generate what type of event? A) ActionEvents B) ChangeEvents C) WindowEvents D) MouseEvents Menus are attached to windows by calling method ____________. A) addMenuBar B) setJMenuBar C) setMenu D) addJMenuBar Solution B) Class JTextField extends JTextComponent; Class JPasswordField extends JTextField. A) ActionEvent, ActionListener B) getPassword B) normally several JRadioButtons are grouped together and are mutually exclusive A) drawing is performed differently on every platform that supports Java. A subclass of Graphics must be created that uses the drawing capabilities of the current platform. A) ActionEvents B) setJMenuBar.

In java languageWhich of the following is falseA) A JPasswordFi.pdf

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in a completely randomized (one-way) anova, with other things being .pdf

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I want you to go the Next Generation Sunshine State standards (http://www.cpalms.org/Public/search/Standard) homepage. Choose a grade level (not K) and look at all the LIFE SCIENCE standards that are found under each of the Big Idea/Supporting Idea topics. To see all the standards, click on the bolded letters. please provide the grade level and copy the standards Summarize what you, as a teacher, would be responsible for teaching your students. Discuss how this framework changed your thoughts on what you, as a teacher, would be responsible for teaching your students for the standards? Discuss how insect topics might be used to help illustrate each of those LIFE SCIENCE standards for that grade level. What insect information do you think could fit with that standard? How could insects be used in related lessons? Solution Summary of grade 7 according to my skills are :- Cell Biology 1. All living organisms are composed of cells, from just one to many trillions, whose details usually are visible only through a microscope. As a basis for understanding this concept: a. Students know cells function similarly in all living organisms. b. Students know the characteristics that distinguish plant cells from animal cells, including chloroplasts and cell walls. c. Students know the nucleus is the repository for genetic information in plant and animal cells. d. Students know that mitochondria liberate energy for the work that cells do and that chloroplasts capture sunlight energy for photosynthesis. e. Students know cells divide to increase their numbers through a process of mitosis, which results in two daughter cells with identical sets of chromosomes. f. Students know that as multicellular organisms develop, their cells differentiate. Genetics 2. A typical cell of any organism contains genetic instructions that specify its traits. Those traits may be modified by environmental influences. As a basis for understanding this concept: a. Students know the differences between the life cycles and reproduction methods of sexual and asexual organisms. b. Students know sexual reproduction produces offspring that inherit half their genes from each parent. c. Students know an inherited trait can be determined by one or more genes. d. Students know plant and animal cells contain many thousands of different genes and typically have two copies of every gene. The two copies (or alleles) of the gene may or may not be identical, and one may be dominant in determining the phenotype while the other is recessive. e. Students know DNA (deoxyribonucleic acid) is the genetic material of living organisms and is located in the chromosomes of each cell. Evolution 3. Biological evolution accounts for the diversity of species developed through gradual processes over many generations. As a basis for understanding this concept: a. Students know both genetic variation and environmental factors are causes of evolution and diversity of organisms. b. Students know the reasoning used by C.

I want you to go the Next Generation Sunshine State standards (http.pdf

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How are macromolecules transported across the nuclear envelope? Describe the structure that regulates nuclear membrane permeability. Solution The transportation of molecules in and out of the nucleus is controlled by the nuclear pore complex(NPC). Small molecules move through the nuclear membrane with ease but for macromolecules like proteins, RNA etc, they require an association with transportation factors like importins, karyopherins and exportins. Proteins that need to be carried into the nucleus have nuclear localization signals (NLS) which bind to the importins. NLS is a sequence of Amino acids which act as a tag. They are mostly hydrophilic but hydrophobic sequences have also been documented. Nuclear Import: Importins bind to the macromolecule to be transported in the cytoplasm itself. The complex then interacts with the NPC and passes through the channel. Once inside the nucleus, it interacts with Ran-GTP and the importin dissociates from the macromolecule it was carrying. Then the importin-Ran-GTP complex is transported to the cytoplasm where it is separated from importin by the Ran Binding Protein (RanBP). After being separated from importin, GTPase activating protein(GAP) binds with Ran-GTP and induces hydrolysis of GTP to GDP. The Ran-GDP produced binds with the nuclear transport factor(NUTF2) and then returns to the nucleoplasm. Inside the nucleus, Ran-GDP interacts with guanine nucleotide exchange factor(GEF) which replaces GDP with GTP and Ran-GTP is born again to repeat the cycle. Nuclear Export: In the nucleoplasm, exportin binds the macromolecule to be transported outside the nucleus and Ran-GTP. The complex diffuses through the pore to the cytoplasm where it dissociates. Ran- GTP binds to GAP and hydrolysis resulting in Ran-GDP. Ran-GDP is returned to the nucleus where it exchanges its ligand for GTP. Nuclear Pore: The nuclear pore complex is made of proteins called nucleoporins. Around 50% of nucleoporins contains solenoid protein domains (alpha or beta-propeller fold) and the rest might have both as separate structural domains. Each NPS has about 456 individual protein molecules with 30 distinct nucleoproteins. These are highly flexible proteins that lack an ordered secondary structure..

How are macromolecules transported across the nuclear envelope Desc.pdf

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GeneticsBackground info About 30 of the Irish population has red.pdf

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Fishes are at the base of the vertebrate Tree of Life and fishes do not have four limbs! However, earlier tetrapods are sister groups of fishes. What fish species are sister to or considered an earlier tetrapod? Solution The evolution of Tetrapods occur 0.4 billion years ago. It occur in Devonian period. The ancestors for tetrapods are Lobe-finned fishes or Sarcoptyrigaens. The living sarcoptygians are Coelacanths and lung fishes.

Fishes are at the base of the vertebrate Tree of Life and fishes do .pdf

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Consider two point sources separated by a distance of 10 meters. The longitudinal waves emitted from the two sources have the wavelength of = lm. The velocity of the wave is 2m/sec. What is the frequency f and the period T? Solution Frequency=Velocity/wavelength =(2m/sec)/1m =2 Hz Time period= 1/Frequency =1/2 =0.5 sec a) Amplitude at C =0.1+0.1=0.2 m (Since waves are in phase) b) Waveform will be at same amplitude that is 0.2 m c) If the distance between the points A and B is changed. The waves interfere distructively. Therefore, amplitude is zero. That is 0.1-0.1=0 d) If source A is delayed by 0.5 sec The path difference is= 2(0.5)=1 m Therefore, waves interfere constructevely. The wave amplitude at point C is., 0.1+0.1=0.2 m.

Consider two point sources separated by a distance of 10 meters. The .pdf

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Consider the following system microprocessor based system: Program the PLD such that interface memory chips at following starting addresses: What is the capacity of each memory chip? What is the MPU memory space? Solution Implementation of SAP SCM Modules for DP, SNP, EIS Regions, US, Canada, Mexico, Chile, Peru, Columbia and Venezuela. Provide technical solution leadership to the Program Director, Project Managers and GSLT Manage IS and vendor stakeholder relationships relevant to the technical solution Ensure technical project and product meet governance, security and CARM guidelines Ensure product is accepted into the support organization Provide status and support to Project Managers in the above activities.

Consider the following system microprocessor based system Program t.pdf

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Develop an algorithm for the 4-sum problem. Solution The algorithm requires creating another array TWOSUMS which has the sum of all possible pair of numbers in the array. We also need to store the index of numbers along with the sum to keep track that the numbers are distinct. Here’s how each entry in the TWOSUMS array will be stored. class Entry{ int sum; int indexa; int indexb; } Entry TWOSUMS[]; Algorithm.

Develop an algorithm for the 4-sum problem.SolutionThe algorit.pdf

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Describe the repair mechanisms used during DNA replication. Solution DNA like any other molecule can undergo a variety of chemical reactions. Because DNA uniquely serves as a permanent copy of the cell genome, however, changes in its structure are of much greater consequence than are alterations in other cell components, i.e RNA’s and Proteins. Mutations can consider the incorporation of incorrect bases during DNA replication. And also, various chemical changes occur in DNA either spontaneously or as a result of exposure to chemicals or radiation. Such damage to DNA can block replication or transcription, and can result in a high frequency of mutations—consequences that are unacceptable from the standpoint of cell reproduction. To maintain the integrity of their genomes, cells have therefore had to evolve mechanisms to repair damaged DNA. DNA repair mechanism can be divided into two general classes: (1) direct reversal of the chemical reaction responsible for DNA damage, and (2) removal of the damaged bases followed by their replacement with newly synthesized DNA. The rate of DNA repair is dependent on many factors, including the cell type, the age of the cell, and the extracellular environment. A cell that has accumulated a large amount of DNA damage, or one that no longer effectively repairs damage incurred to its DNA, can enter one of three possible states: 1. an irreversible state of dormancy, known as senescence 2. cell suicide, also known as apoptosis or programmed cell death 3. unregulated cell division, which can lead to the formation of a tumor that is cancerous The DNA repair ability of a cell is vital to the integrity of its genome and thus to the normal functionality of that organism. Direct reversal Cells are known to eliminate three types of damage to their DNA by chemically reversing it. These mechanisms do not require a template,the types of damage they counteract can occur in only one of the four bases. Such direct reversal mechanisms are specific to the type of damage incurred and do not involve breakage of the phosphodiester backbone. The formation of pyrimidine dimers upon irradiation with UV light results in an abnormal covalent bond between adjacent pyrimidine bases. The photo reactivation process directly reverses this damage by the action of the enzyme photolyase, whose activation is obligately dependent on energy absorbed from blue/UV light (300–500 nm wavelength) to promote catalysis. The second type of damage, methylation of guanine bases, is directly reversed by the protein methyl guanine methyl transferase (MGMT), The third type of DNA damage reversed by cells is certain methylation of the bases cytosine and adenine. Excision Repair mechanisms Single strand damage: When only one of the two strands of a double helix has a defect, the other strand can be used as a template to guide the correction of the damaged strand. In order to repair damage to one of the two paired molecules of DNA, there exist a number of excis.

Describe the repair mechanisms used during DNA replication.Soluti.pdf

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Describe the current state of the human beta globin gene family and the processes that were involved in generating it from a single ancestral globin gene. How do the members of the gene family differ and how do these differences relate to their expression and function? How does the human beta globin gene family differ from the beta globin gene families in other vertebrates? What do these differences imply about the evolution of this gene family in different species; are the same processes that shaped the human gene family also at work in the other species and if so why are the gene families different in the other species? What differences in the beta globin genes are found among the current human population and what are the genetic consequences of these differences? Solution human -globin locus are composed of five gene, that are located on short chromosome number 11, This locus also contains the delta, gamma-A, gamma-G, and epsilon globin. Expression of all genes are controlled by a single locus control region, and all beta genes are differentially expressed during development. Duplication + divergence of ancestral gene is the major process involve in generation of beta gene form single ancestral globin gene. Human beta globulin gene come under gene family because of gene family is set of various similar gene that form by duplication of single original Gene & almost similar biochemical function. Gene family member are different in several way 1. arrangement of genes directly reflect to differentiation of their the expression during development. For proper Functioning required another alpha globin for complete function for example- Hemoglobin protein is made of two alpha and two beta polypeptide are encoded by different gene. Globin genes are similar in structure Each gene encoded similar but different protein. 2. Expression of gene is regulated in the embryonic erythropoiesis by the many transcription factors like [KLF1] and that which is associated with upregulation of the adult and another [KLF2] which are vital to expression of the embryonic hemoglobin. beta-globin gene are clustered & work as unit of gene regulation, embryo --> express epsilon, fetal --> express gamma, adult --> express beta (and delta) Gene order-position-location- important for regulation of gene. the human beta globin gen family different form beta globin gene family in vertebrate- 1.Polyploidy occurs more often in plant genome than in animal one or at least vertebrate genomes. 2. loss of intron- first & third intron only two intron found in vertebrate hemoglobin & myoglobin genes. Thecentral intron was lost before the divergence of annelids & arthropods and absent in all the vertebrate hemoglobin and myoglobin genes. Intron loss has in different organisms & there is extreme example of the total intron loss in arthropod Chironomus. 3. major gene duplications in vertebrate globin’s. evolution of this gen family in difference species- Gene families are the groups of the.

Describe the current state of the human beta globin gene family and .pdf

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Consider the (Mining dataset shown in the following table The last attribute Cheat is the class attribute. For a three-way -split of these data using attribute Marital Status, compute the information gain using the entropy function fix measuring impurity. Consider that the three-way (multi-way) spin a does by using the three possible values Single. Married, and Divorced of the .Marital Status attribute Solution Info(D) = -MI=1pi log2(pi); Pi= |Ci,D|/|D| InfoA(D) = vj=1 |Dj|/ |D| *Info(Dj): Gain(A) = Info(D)-InfoA(D): Info(D)=- (3/10) log2 (3/10) – (7/10) log2 (7/10) =0.522+0.361 =0.883 Info Marital status (D) = (4/10)* (-(2/4) log2(2/4) - (2/4) log2(2/4)) + (4/10)*(-(0/4) log2(0/2)- (4/4) log2(4/4) +(2/10)*(-(1/2) log2(1/2) -(1/2) log2(1/2)) = 0.4 + 0 + 0.2 = 0.6 Therefore Gain (Marital Status) = 0.883-0.6 = 0.283.

Consider the (Mining dataset shown in the following table The las.pdf

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