Prove or disprove A non-constant analytic function can have at most .pdf

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Prove or disprove: A non-constant analytic function can have at most finitely many zeros on a closed disk. Solution Suppose an analytic function f has infinitely many zeros on some closed disk D. Then there exists a sequence of zeros in D with a limit point in D. Thus by the identity theorem (Let D be a domain and f analytic in D. If the set of zeros Z(f) has a limit point in D, then f 0 in D.), f is identically zero and thus constant. so for a non-constant analytic function must have finite zeros on a closed disk.

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Explain microscope in simplest terms Definition and significance of contrast, absorption, refraction, and scattering to light microscopy I know what contrast is, but what is the significance of absorption, refraction, and scattering of the light to a light microscope? When do all of these come in to play Solution Microscope- Microscope is optical instrument, in which light makes image clear to see microorganism, plant and animal cell etc. Significance of absorbance, scattering and reflection in microscope- In light microscopy abortion play the role to see the object in term to see the colour when visible light (VIBGYOR) put on to object the all waves are absorbed but a particular wave length is reflect by this wave we can see the colour of the object. For example the leaf of the plant look green because all wave are absorbed but green wave length is reflect. In scattering of light microscope the light is scattered by rough or rigid area of the object, from this we obtain size and shape of the object. When put the visible light on object it absorbed, scattered and reflect by ht object at a time by absorbance and reflection we see the colour and by scattering see the shape and size of the object.

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Find the conditional probability of the indicated event when two fai.pdf

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eukaryotic cell or prokaryotic cell Solution Prokaryotic cells are devoid of an organized membrane bound nucleus and different cell organelles. Incontrast a Eucaryotic cells have an organized membrane bound nucleus and different cell organelles. Based on this information: The images 1, 2 and 3 are of Eucaryotic cells While the image 4 which is having no membranous organelles inside the cell is of procaryotic cell.

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Describe how to make 10ml of a 1:25 dilution from a 1 times 10^12 cells/ml stock. What is the resulting cell concentration? Solution = 1/25 Step 1 Take one part stock and add 4 parts water to get 1/5 dilution i.e. 1x 10 12 cells/ml/ 5 = 0.2 1x 10 12 cells/ml Step 2 Now take one part stock of this cell suspension and add 4 parts water to get 1/5 dilution leading to 1/25 dilution. = 0.2 x 10 12 cells/ml/ 5 0.04 xx 10 12 cells/ml the total volume will be 10 ml b) = 4 x 10 10 cells/ml this is the final cell concentration..

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Compare organelles in both plant and animal cells. Identify the structures that are unique to each type of cell? HINT: There are ONLY four (4) unique structures for plant cells and ONLY two (2) unique structures for animal cells. Solution Animal cells Plant cells 1). Do not contain cell wall Contains cell wall 2). Shape of the cell can be changed Shape of the cell cannot be changed 3). Lysosomes are present Lysosomes are rarely observed 4). Small vacuoles are present One single large vacuole is present 5). Animal cells divide by pinching off the cytoplasm, The plant cells form cell plate during the division. 6). Tight cell junctions can be formed between the cells. No tight cell junctions can be formed. 7). Chloroplasts absent Cloroplasts present 8). Can undergo both mitosis and meiosis Can undergo both mitosis and meiosis Animal cells Plant cells 1). Do not contain cell wall Contains cell wall 2). Shape of the cell can be changed Shape of the cell cannot be changed 3). Lysosomes are present Lysosomes are rarely observed 4). Small vacuoles are present One single large vacuole is present 5). Animal cells divide by pinching off the cytoplasm, The plant cells form cell plate during the division. 6). Tight cell junctions can be formed between the cells. No tight cell junctions can be formed. 7). Chloroplasts absent Cloroplasts present 8). Can undergo both mitosis and meiosis Can undergo both mitosis and meiosis.

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Soap molecules have a hydrophilic head and a hydrophobic tail. What happens when you put soap in water? The hydrophobic tails cluster together and exclude water The hydrophilic heads interact with water The hydrophobic tails bind to one another Only A and B are true A, B, and C are all true Soap molecules have a hydrophilic head and a hydrophobic tail. What happens when you put soap in water?A. The hydrophobic tails cluster together and exclude waterB. The hydrophilic heads interact with waterC. The hydrophobic tails bind to one anotherD. Only A and B are trueE. A, B, and C are all true Solution Soap molecules are amphipathic in nature, i.e. they have both polar and non-polar constituents in them. The polar head, composed of a salt group, is joined to a hydrophobic tail, which is non- polar. When the soap molecules are placed in water, these molecules orient themselves such that the polar head group interacts with water, whereas the hydrophobic chains bind to one another, so as to minimize the contact with water. The hydrophobic chains form a cluster, and this structure is known as micelles. Thus, option (E) is correct, because statements A, B, and C are all true..

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What are TEs? Comment on Are there any distinctive features Why might biologists want to study what do you know different kinds associated with TEs? What are TES? about transposable What types of organisms contain these? elements (TEs) in TEs? general? Vague (1); OK (1.5); Good (3) Vague (1), OK (1.5), Good (3) Vague (1): OK (1.5); Good (3 How might P-elements have How do P-elements move around Why are Pelements important for entered Drosophila melanogaster? the genome? Connect your ideas biologists and Drosophila What do you know What do you know about P-element to the special P-elements used in researchers in particular? about P-elements structure? How big are the native this lab. What is the connection What do you know about the Gene in Drosophila? elements? between transposition of mobile Disruption Project (GDP) in DNA elements and DNA Drosophila? replication? Vague (1): OK (1.5): Good (3 Vague (1): OK (2): Good (4 Vague (1): OK (2): Good (4 Solution 1.Transposable elements or transposons are defined as the DNA sequence that are able to move from one location to the other in the genome.They were first discovered by Barbara McClintock in maize and are also called as the \"jumping genes\". Transposons are found in almost all organisms starting from bacteria to higher organisms i.e from prokaryote to eukaryotes and are known to have played a significant role in evolution. There are two major types of transposons .they are: Class I transposons and Class II transposons Class I transposons are called as retrotransposons as they first transcribe the DNA into RNA and then use the enzyme reverse transcriptase to make a copy of the DNA from RNA and insert it in a new location. Class II operate by \"Cut and Paste\" mechanism and contain DNA that moves directly from place to place.They are cut at their location and are inserted into a new location.Distinctive features of transposons; 1.One of the most defining property of a transposon is its mobility. 2.They are diverse in nature some move by DNA intermediates (class I) while some move by RNA intermediates(ClassII) 3.Due to their ability to rearrangement they play a significant role in evolution. 4.some transposons activate or inactivate the genes depending on their position of insertion.egEnhancement of oncogeneic activity. Transposons cause deletions and inversions..

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Which myth of motivation is the most important Are there other myth.pdf

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What is an interface? How is extending a class different from implementing an interface? Solution Inteface: Interface is nothing but a class which contains abstract methods and constants only. Abstract methods are the methods which have declaration but no bosy implementation. Interfaces can not have constrcutor and can not be instanstiated. By using interface we can achive multiple inheritance. Inheritance goes with \"implements\" keyword in interfaces where as Inheritance goes with \"extends\" keyword in class. If we implements interface in sub class then we should implement all abstract methds in interface in sub classses. interface syntax public interface interfacename{ public abstract void abstractMethod1(); public abstract void abstractMethod2(); public static final CONSTACT = 1000; } Class syntax public class className{ public void methodA(){ } public void methodB(){ } } Inheritance goes with implements in interface public class A implements interfaceName{ } Inheritance goes with extends in class public class A extends ClassName{ } Inheritance goes with implements in interface public class A implements interfaceName{ }.

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SAY. Write your answer in the space provided or on a separate sheet of pap In the fruit fly. Drarephila melanu raster, a spineless (no wing a male that is claret bristles) female ny mated to (dark eyes) and hairless (no th otypically wild-type F1 f nai progeny were mated to fully homozygous (mutant) males, and the following progeny (1000 total) were observed, claret, spineless 18 309 claret, hairless hairless, claret, spineless 140 hairless hairless spineless (a) Which gene is in the middle? (b) With respect to the three genes mentioned in the problem, what are the genotypes of the homozygous Parents used in making the wild F1 heteroaygote? (c What are the map distances between the three genes? (d) What is the coefficient of coincidence? Solution Answer: (1) (a) Spineless = (s + +) [Parental] Claret hairless = (+ c h) [Parental] Claret = =(+ c +) [double cross over] hairless,spineless = (s + h) [double cross over] Therefore, Parentals = s + + , + c h and Double crossovers = + c + , s + h Analysing, we get: s+ s+, +c +c and h+ +h. h is different, so, h (hairless) is in the middle. (b) The genotypes of the homozygous parents are = s + + / s + + and + c h / + c h (c) The map distances have been calculated as under: (Spineless – hairless) = (Number of recombinants)/1000) * 100 = (32 + 38 + 12 + 18)/(1000) * 100 = (100/1000) * 100 = 10% = 10 map units (mu) (hairless-claret) = (140 + 130 + 12 + 18)/1000) * 100 = (300/1000) * 100 = 30% = 30 map units (mu) (d) Coefficient of coincidence = observed double crossovers/expected double crossovers So, we need to calculate expected double cross overs: Expected double cross overs = recombination frequency in region 1 (map units / 100) X recombination frequency in region 2 (map units / 100) = .(10/100) * (30/100) = 0.1 * 0.3 = 0.03 Therefore, Coefficient of coincidence = [(18 + 12)/1000] / 0.03 = (30/1000) / 0.03 = 0.03 / 0.03 = 1.

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What is validity? What is reliability? Discuss the relationship can a measure be valid, but not reliable? Reliable but not valid? Explain. Solution ; The term validity is defined as the accuracy of the interpretations which we can make out from the test scores On the other hand Reliability is defined as the the consistency or stability of the test scores No a measure cannot be valid unless it is reliable.It has to be reliable first in order to be valid .On the other hand the measure can be reliable but not valid.The measure can be consistent without any accuracy..

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With a VPN ______ a. users are often provided a token that displays a constantly changing password c. users can gain wide area network access b. users can gain access to an organization’s private network d. all of the above Solution With a VPN users are frequently provided a token that displays a constantly changing password, users can gain access to an organization’s private network and users can gain wide area network access. Answer: d. all the above.

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Which of the following is the basis of genes that perform different but related functions in evolutionary related organisms? Conversion of exons into introns Gene duplication and subsequent mutation Whole genome duplications the presence of a nucleus in prokaryotes Solution C) whole-genome duplications The convergence clarification is implausible for adjacent associated sequences, such as resemblance but still arrogate similar structures, suppress identical or associated amino acid . The evolutionary starting point, of such conservation of purpose among orthologs. These incorporate dissimilar forms of genome rearrangement, genome duplication,.

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What are three possible methods that a eukaryote might ultilize to allow a single DNA sequence to code for at least a portion of more than one protein? Solution There are more than one way that a specific segment of a DNA, called genes, can encode multiple protein types 1. Alternative splicing is a process where exons or portions of exons or noncoding regions within a pre-mRNA transcript are differentially joined or skipped, resulting in multiple protein isoforms being encoded by a single gene. 2. The secend way to of this is use of alternative initiation or termination codons which allows two proteins to be generated where one is equivalent to a fragment of the other. 3. the third way of this is Overlapping genes which occur in the relatively simple situation in which one gene is part of the other. The first half (or second half) of a gene is used independently to specify a protein that represents the first (or second) half of the protein specified by the full gene..

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The color of the RBC\'s cytoplasm is (?) and it has a biconcave shape because of the lack of a (?). One of its functions, is to (?)(?). (?) = fill in the blank Solution The color of the RBC\'s cytoplasm is Red colour The cytoplasm of erythrocytes is rich in hemoglobin, an iron-containing biomolecule that can bind oxygen and is responsible for the red color of the cells t has a biconcave shape because of the lack of a nucleus. biconcave shape gives them a vastly greater surface area than a spherical cell of similar volume, which allows them to absorb oxygen more efficiently. Red blood cells achieve this shape by losing their nucleus and many other organelles during development. One of its functions, is to transport of O2.

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/** The business logic of a Parking Pay Station. Responsibilities: 1) Accept payment; 2) Calculate parking time based on payment; 3) Know earning, parking time bought; 4) Issue receipts; 5) Handle buy and cancel events. This source code is from the book \"Flexible, Reliable Software: Using Patterns and Agile Development\" published 2010 by CRC Press. Author: Henrik B Christensen Department of Computer Science Aarhus University Please visit http://www.baerbak.com/ for further information. Licensed under the Apache License, Version 2.0 (the \"License\"); you may not use this file except in compliance with the License. You may obtain a copy of the License at http://www.apache.org/licenses/LICENSE-2.0 Unless required by applicable law or agreed to in writing, software distributed under the License is distributed on an \"AS IS\" BASIS, WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. See the License for the specific language governing permissions and limitations under the License. */ public interface PayStation { /** * Insert coin into the pay station and adjust state accordingly. * @param coinValue is an integer value representing the coin in * cent. That is, a quarter is coinValue=25, etc. * @throws IllegalCoinException in case coinValue is not * a valid coin value */ public void addPayment( int coinValue ) throws IllegalCoinException; /** * Read the machine\'s display. The display shows a numerical * description of the amount of parking time accumulated so far * based on inserted payment. * @return the number to display on the pay station display */ public int readDisplay(); public int readDisplay2(); /** * Buy parking time. Terminate the ongoing transaction and * return a parking receipt. A non-null object is always returned. * @return a valid parking receipt object. */ public Receipt buy(); /** * Cancel the present transaction. Resets the machine for a new * transaction. */ public void cancel(); } /** Implementation of the pay station. Responsibilities: 1) Accept payment; 2) Calculate parking time based on payment; 3) Know earning, parking time bought; 4) Issue receipts; 5) Handle buy and cancel events. This source code is from the book \"Flexible, Reliable Software: Using Patterns and Agile Development\" published 2010 by CRC Press. Author: Henrik B Christensen Department of Computer Science Aarhus University Please visit http://www.baerbak.com/ for further information. Licensed under the Apache License, Version 2.0 (the \"License\"); you may not use this file except in compliance with the License. You may obtain a copy of the License at http://www.apache.org/licenses/LICENSE-2.0 Unless required by applicable law or agreed to in writing, software distributed under the License is distributed on an \"AS IS\" BASIS, WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. See the License for the specific language governing permissions and limitations under the License. */ public class PayStationImpl imple.

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Refer to Exercise 2.25. The agreement between the percentages for the randomized study and the actual population percentages for various groups of people may allow us to have more confidence in the main results of the survey, which are the percentages of various HIV-related risk groups as shown next. a. For the nation as a whole, what percentage of those between the ages of 18 and 75 are at risk HIV infection? What is the percentage for the high -risk cities? (Actually, these percentages go much higher if sexual practices for more than the past year are taken into account.) b. The percentages can be used as estimates of probabilities for anticipating what might happen in certain situations. For example, if a national firm is to hire 1000 workers across the country, how many, would they expect to be at risk for HIV infection? What happens to this expected value if the firm hires all 1000 workers in high -risk cities? Solution BY RISK GROUP Gay, bisexual, and other men who have sex with men (MSM) of all races and ethnicities remain the population most profoundly affected by HIV. In 2010, the estimated number of new HIV infections among MSM was 29,800, a significant 12% increase from the 26,700 new infections among MSM in 20082. Although MSM represent about 4% of the male population in the United States4, in 2010, MSM accounted for 78% of new HIV infections among males and 63% of all new infections2. MSM accounted for 54% of all people living with HIV infection in 2011, the most recent year these data are available1. In 2010, white MSM continued to account for the largest number of new HIV infections (11,200), by transmission category, followed closely by black MSM (10,600)2. The estimated number of new HIV infections was greatest among MSM in the youngest age group. In 2010, the greatest number of new HIV infections (4,800) among MSM occurred in young black/African American MSM aged 13.

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