Chromosome Structure Explain why each chromosome in a eukaryotic cel.pdf

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Chromosome Structure: Explain why each chromosome in a eukaryotic cell has multiple origins of replication but one and only one centromere. Solution Answer: Eukaryotic genome is organised in linear chromosomes which are separate (not linked). Since the eukaryotic DNA is large, it could be way too time consuming to have only 1 origin of replication. Multiple chromosomes and large genome size brings in multiple origin of replication to quicken the process of replication. Multiple origins of replication allow many parts of the DNA to be replicated simultaneously..

Chromosome Structure: Explain why each chromosome in a eukaryotic cell has multiple origins
of replication but one and only one centromere.
Solution
Answer:
Eukaryotic genome is organised in linear chromosomes which are separate (not linked). Since
the eukaryotic DNA is large, it could be way too time consuming to have only 1 origin of
replication. Multiple chromosomes and large genome size brings in multiple origin of replication
to quicken the process of replication. Multiple origins of replication allow many parts of the
DNA to be replicated simultaneously.

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Verified the following trig identities and show all work. csc4 t-cot4 t = csc2t+cot2 t cosB/1-sinB = secB+tanB 1/csc y-cot y = csc y + cot y Cot 4u-1/cot 4u+1 = 1-tan 4u/1+tan 4u (sec t + tan t)2 = 1+sin t / 1-sin t Cos3x –sin3x /cos x-sin x = 1+sin x cos x Csc x /1+csc x – csc x/ 1-csc x= 2sec2 x (1-tan2 theta)2 = sec4 theta – 4 tan2 theta Solution 1)csc4t-cot4t a2-b2=(a-b)(a+b) =(csc2t-cot2t)(csc2t+cot2t) =1*(csc2t+cot2t) =csc2t+cot2t 2)cosB/1-sinB multiply and divide by (1+sinB) =cosB(1+sinB)/((1-sinB)(1+sinB)) =cosB(1+sinB)/(1-sin2B) =cosB(1+sinB)/(cos2B) =(1+sinB)/cosB =(1/cosB)+(sinB/cosB) =secB+tanB.

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Toxaphene is an insecticide that has been identified as a pollutant in the Great Lakes ecosystem. To investigate the effect of toxaphene exposure on animals, groups of rats were given toxaphene in their diet. The article “Reproduction Study of Toxaphene in the Rat” (J. of Environ. Sci. Health, 1988: 101-126) reports weight gains (in grams) for rats given a low dose (4 ppm) and for control rats whose diet did not include the insecticide. The sample standard deviation for 23 female control rats was 32 g and for 20 female low-dose rats was 54 g. Does this data suggest that there is more variability in low-dose weight gains than in control weight gains? Assuming normality, carry out a test of hypotheses at significance level .05 For this problem, make a two-sided t interval & perform an F test for the ratio of variance, & create a confidence interval for the ratio of the variance. Solution.

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The number of passengers (including the driver) in cars that cross a certain bridge is a random variable Y having mean E(Y) = 2.7 and variance V(Y) = 1.2. The toll on the bridge is $3.00 per car plus $0.50 per person in the car. The mean and variance of the total amount of money that is collected from a car that crosses the bridge are: (a) mean = $8.60, variance = $.30 (b) mean = $4.35, variance = $3.30 (c) mean = $8.60, variance = $.60 (d) mean = $4.35, variance = $.30 (e) mean = $1.35, variance = $.30 Solution The amount they pay in terms of Y is A = 3.00 + 0.50Y Thus, E(A) = E(3.00 + 0.50Y) = 3.00 + 0.50E(Y) = 3.00 + 0.50*2.7 E(A) = 4.35 Also, Var(A) = Var(3.00 + 0.50Y) = 0.50^2 Var(Y) = 0.50^2 * 1.2 Var(A) = 0.30 Thus, it is OPTION D. [ANSWER, D].

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the projected number of households( in millions) with digital video records thats allow viewers to record shows onto a server and skip commercials are in the following 5. The projected number of households (in millions) with digital video recorders that allow viewers to record shows onto a server and skip commercials are given in the following table: Year Households 31.2 a) Let t = 0 correspond to the beginning of 2006, and use QuadReg to find a second- 2010 130.2 2006 2008 71.6 2009 97.0 2007 49.0 degree polynomial regression model based on the given data. (2 pts) b) Using your regression equation, find the number of households in 2008. (2 pts) c) Using your regression equation, find the number of households in 2013. (2 pts) Solution a) Using technology to get the best quadratic fit, y^ = 2.4t^2 + 15t + 31.4 [answer] where y^ = predicted number of housholds. b) Plugging in t = 2 to y^ above, y^ = 2.4(2)^2 + 15(2) + 31.4 = 71 [answer] c) Plugging in y = 7 above, y^ = 2.4(7)^2 + 15(7) + 31.4 = 254 [answer].

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The plant pathogen Xylella fastidiosa is a gram negative that attaches to surfaces and forms a biofilm in host plants. Which of the following typically triggers an immune response? a) Pili b) Teichoic acid c) LPS d) EPS Solution Ans--(C) : LPS Lipopolysaccharide (LPS) is the predominant macromolecule displayed on the cell surface of Gram-negative bacteria and thus mediates bacterial interactions with the environment and potential hosts Utilizing a mutant that produces truncated O antigen we present results that link this LPS structural alteration to a significant decrease in the attachment of X,Xylella fastidiosa is a plant pathogenic bacterium that lives inside the host xylem vessels where it forms biofilm believed to be responsible for disrupting the passage of water and nutrients..

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a) What is the Chromosome Theory of Inheritance and how was it determined? (SHOW WORK & EXPLAIN) b) Describe the contributions of Sutton, Boveri, McClung, Stevens, Wilson and Morgan to determining that genetic information is carried on chromosomes. (SHOW WORK & EXPLAIN) c) Describe the XX-XO, XX-XY and ZZ-ZW mechanisms of sex determination and give examples of organisms that use each mechanism. (SHOW WORK & EXPLAIN) d) How does the environment affect sexual phenotype in turtles and alligators? (SHOW WORK & EXPLAIN) Solution a) Chromosomal Theory of Inheritance is a fundamental theory of genetics.Thomas Hunt Morgan establishes the Chromosomal theory of heredity.In 1907 Morgan was based on experiments by using Drosphila melanogaster or fruit flies. Drosphila have red eyes.Red eye colour is dominant.Morgan discovered a recessive mutation that caused white eyes. B) Boveri-Sutton chromosome theory identifies chromosomes as the carriers of genetic material which states that chromosomes carry the unit of heredity. McClung : A zoologist studied the behaviour of chromosomes in the sex cells and discovered the sex determining chromosome. Stevens: Stevens was interested in the process of sex determination. Morgan proposed the Chromosomal Theory of Heredity by using a common fruit fly. C) Sex determination is a biological system that determines the development of sexual characteristics. XX-XO types : In this type females have two homomorphic sex chromosomes XX ,male have one sex chromosome XO. This type of sex determination seen in cockroaches. XX-XY : This sex determination occurs in human beings.Females have two identical homologous sex chromosome XX and males have XY. ZZ-ZW : Females carry two types of sex chromosome ZW heterogametic and males carry ZZ homogametic. This type of sex determination occurs in fishes reptiles. D) Sex determination in reptiles are classified into two categories that is genotypic sex determination and temperature dependent. Environmental sex determination in which temperature is a crucial factor in embryonic development. In many turtle species eggs from cooler nests hatch males and from warmer nests hatch the females..

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Prove that the chromatic number of a cycle graph_n of odd length equals 3. Solution Chromatic Number: The chromatic number of a graph G is the least k for which a k-coloring of G exists. First note that the chromatic number must be at least 2 for any graph which has an edge in it, including all cycles. We now prove the theorem by induction on n. We will consider two base cases, C3 and C4. C3 is isomorphic to K3 which has chromatic number 3. C4 can be colored with two colors by giving opposing corners of the square the same color. For n > 4, we can take a coloring of Cn-2 and insert 2 adjacent vertices and edges and then color the new vertices appropriately to get a coloring of Cn. Thus the chromatic number of Cn is not greater than that of Cn-2. In the case where n is odd, note that if Cn had chromatic number 2, we could remove two adjacent vertices and edges to get a 2-coloring of Cn-2 which contradicts the inductive hypothesis since n-2 must be odd if n is odd. Therefore for odd numbers the chromatic number will be 2+1 = 3..

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Many non-scientists have focused their criticism of evolution on the fact that even scientists have criticized Darwin’s theory. Is this true? In what respect did Darwin get it wrong, or maybe in what way has Darwin’s theory been modified? Solution The \"Origin of Species\" of Darwin described about the heritable change in an organism over a period of time due to natural selection. Scientists also criticized of Danwin\'s opinion on use and disuse theory. At the time of Origin of species, Darwin do not know about the existed of microorganisms and his theory of tree of life became wrong. Pangenesis theory of Darwin also became wrong as he do not know anything about genetics..

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iQ2 In a study about children?s pocket money, two samples of schoolchildren were asked how much money they received each week. In one sample, of size 80, the mean is ?6.65. In the other sample, of size 41, the mean is pound 8.21. Calculate the mean based on the combined sample. Give your answer correct to two decimal places. (Do NOT include the pound sign in your answer.) Solution n1=80 x1=6.65 n2=41 x2=8.21 combined mean = (n1*x1+n2*x2)/( n1+n2) =( 80*6.65+41*8.21) /( 80+41) =868.61/121 =7.1785 Combined mean =7.18 ( two decimals).

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A population of 10,000 students spends an average of $8.50 a day on dinner. The standard deviation of the expenditure is $4. A simple random sample of 100 students is taken. What is the standard deviation of the sampling distribution of the sample mean? Round your answer to 3 decimal places and no dollar sign. Solution As s(X) = s/sqrt(n) Then, as s = 4, n = 100, s(X) = 4/sqrt(100) = 0.4 [ANSWER].

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a local fan club surveyed 67 of its members regarding their viewing habits last weekend, and the 44 following information was obtained: 30 members watched an episode of Naruto, 44 watched an episode of Death Note, 23 watched an episode of Inuyasha, 20 watched both Naruto and Inuyasha, 5 watched Naruto and Inuyasha but not Death Note, 15 watched both Death Note and Inuyasha, and 23 watched onlyDeath Note. Draw a Venn Diagram. Solution.

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A gas is a fluid. It has less molecular space and freedom of motion than liquids as well negligible cohesive intermolecular forces. Shear thinning and thickening are types of non- Newtonian fluids. Gage pressure = Absolute Pressure + 14.7 4. The specific weight of an unknown object is 75 lbf/ft^3. The object is placed in water. The object will float within the water but not float to the water surface. The center of mass is the point through which the resultant force acts on a submerged inclined plane. Solution 1.TRUE...ie correct explaination 2.TRUE SHEAR THICKENING IS DILATENT...SHEAR THINING IS PSEUDOPLASTIC 3.FALSE....ABSOLUTE PRESSURE=GAUGE+ ABSOLUTE PRESSURE 4.FALSE..IT IS ALWAYS FLOAT ON THE WATER 5. TRUE...

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A man accidentally let his parakeet out the door. The bird flew up to a power line and sat there. Unable to coax the bird down, the man stands on an aluminium ladder and reaches for the bird with an aluminium pole. Is there any potential danger in this situation? If so, what, and if not, why not? (Figure adapted from: Irwin, J. D. & Kerns, D. v. (1995). Introduction to Electrical Engineering. Upper Saddle River, NJ: Prentice-Hall.) Solution Yes there a potentially fatal scenario because here the man may provide a path to ground from the power line through his body. Since aluminium pole is a conductor so as soon as the man touches power line there there will be a electrical path produced through him and power passes from pole to man and then to ground..

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A culture of yeast was grown in a media containing sucrose labeled by 14C at 2nd position of fructose. Determine the localization of the label in palmitic acid synthesized by the yeast. Solution GLYCOLYSIS result in synthesis of acetyl CoA acetyl CoA formed from fructose will be labeled by 14C as 2nd position of fructose is labeled by 14C this will result in sub cellular localisation of pamitic acid synthesis......by acetyl CoA which is labeled by 14C.

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5. When a phosphate group is found in water: a) The partially-negative oxygen atoms of the water molecules surround the oxygen atoms of the phosphate group. b) The partially-positive hydrogen atoms of the water molecules surround the oxygen atoms of the phosphate group. c) The partially-positive oxygen atoms of the water molecules surround the oxygen atoms of the phosphate group. d) Hydrophobic forces push the phosphate groups together to minimize their disruption on hydrogen bonding between water molecules. e) There is more than one correct answer above. Solution ANS: b) The partially-positive hydrogen atoms of the water molecules surround the oxygen atoms of the phosphate group. Explanation: In the water molecule two polar covalent bonds are present. Among this oxygen molecule had a partial negative charge and both hydrogen atoms partial positive charge. The positive and negative ends of the bond cause charge separation or a dipole. So water molecule had 2 polar covalent bonds and 2 bond dipoles. In phosphate group two oxygen molecules had partial negative charges, so the partially-positive hydrogen atoms of the water molecules surround the oxygen atoms of the phosphate group..

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HMCS Vancouver Pre-Deployment Brief - May 2024 (Web Version).pptx

HMCS Vancouver Pre-Deployment Brief - May 2024 (Web Version).pptx

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Mehran University Newsletter Vol-X, Issue-I, 2024

Mehran University Newsletter Vol-X, Issue-I, 2024

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Sociology 101 Demonstration of Learning Exhibit

Sociology 101 Demonstration of Learning Exhibit

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ICT role in 21st century education and it's challenges.

ICT role in 21st century education and it's challenges.

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Google Gemini An AI Revolution in Education.pptx

Google Gemini An AI Revolution in Education.pptx

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Key note speaker Neum_Admir Softic_ENG.pdf

Key note speaker Neum_Admir Softic_ENG.pdf

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Interdisciplinary_Insights_Data_Collection_Methods.pptx

Interdisciplinary_Insights_Data_Collection_Methods.pptx

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Graduate Outcomes Presentation Slides - English

Graduate Outcomes Presentation Slides - English

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On National Teacher Day, meet the 2024-25 Kenan Fellows

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1. Chromosome Structure: Explain why each chromosome in a eukaryotic cell has multiple origins of replication but one and only one centromere. Solution Answer: Eukaryotic genome is organised in linear chromosomes which are separate (not linked). Since the eukaryotic DNA is large, it could be way too time consuming to have only 1 origin of replication. Multiple chromosomes and large genome size brings in multiple origin of replication to quicken the process of replication. Multiple origins of replication allow many parts of the DNA to be replicated simultaneously.