Which of the following items are stocks, as opposed to flowsQuest.pdf

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Which of the following IP address groups can be used for assignment to computers using the public Internet? A. 10.0.0.0–10.255.255.255 B. 12.222.0.0–12.222.255.255 C. 172.16.0.0–172.31.255.255 D. 192.168.0.0–192.168.255.255 Solution Answer B is the correct answer. It is the only IP address shown in the four options that is a routable IP address. The address groups listed in answers A, C, and D are reserved for private networks and are not routed by the Internet, so those answers are incorrect. You will need to memorize the address groups reserved for private IP networks..

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Which of the following items are stocks, as opposed to flows?
Question options (One or more answers may be correct):
A Loan
A Cat
An invoice
Paid-In Capital
A Loan Payment
Interest Expense
Shareholder Dividends
A Loan
A Cat
An invoice
Paid-In Capital
A Loan Payment
Interest Expense
Shareholder Dividends
Solution
Ans:
Paid-In Capital
Shareholder Dividends

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Blair, R. B. 1996. Land use and avian species diversity along an urban gradient. Ecological Applications 6:506–519. http://academic.uprm.edu/~jchinea/cursos/comunidades/refs/blair1996.pdf Questions 1. The author predicts that different bird species will respond differently to urbanization. What are the three response groups the author expects to find and how is each predicted to respond to urbanization? 2. Although not explicitly stated in the text, the author compared bird densities among sites using analysis of variance (ANOVA). Describe the experimental design of the ANOVA: what were the response and predictor variables, and treatment levels? Explain how the author used replication in both space and time. 3. The author presents data on bird densities within sites in both Table 1 and Figure 3. Which presentation do you find more interpretable and why? What information is common to both, what is unique to either one or the other? Could you limit the redundancy of these two presentations by combining them into one table or figure? If so, which would you choose and how would you lay it out? 4. The golf course had the highest bird species richness, density, and biomass (when Rock Doves were excluded from the analysis), and the second highest diversity (Figure 4). Summarize the author’s explanation for this result. 5. Based on the study results, briefly argue the pros and cons of golf courses as they relate to bird diversity and abundance. Solution 1: The author predicts three groups of birds which he called as urban exploiters, urban avoiders and suburban adaptable. The urban exploiters are expected to get adapted to the changes and to reach their highest densities at the developed sites. The urban avoiders are expected to respond in a sensitive way to the anthropogenic changes in their habitat and are expected to have their maximum densities at the natural sites only. The suburban adaptables are expected to exploit some additional available resources leading to their moderate development..

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Consider a relation T with six attributes ABCDEF where AB is a composite primary key for T. Assume T has the following functional dependencies: AB -> CDEF A -> CEF F -> E In what normal form is relation T? Explain why? Is it possible to decompose T into a number of relations in order to achieve a higher normal form? Explain how. Solution T(A,B,C,D,E,F) AB -> CDEF A -> CEF F -> E The relation T is in First Normal Form assuming all attribute values are atomic. But it is not in second normal form because attribute E is functionally dependent on the attribute F.But F is not part of key attributes so relation is not in second normal form. attribute.A relation is in second normal form if all non key attributes are fully functionally deprndent on attribute keys. Yes , it is possisble to decompose the relation into number of relations in order to achieve the higher normalization. Following are the tables into which we can decompose the relation T Table1(A,C,E,F) Table2 (A,B,D) Table3 (F,E) underlined attributes are primary keys.

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An enzyme aggase requires >16 units of activity for wild type function. Two alleles of aggase have been identified. Wild type allele A_1 produces 12 units of enzyme and mutant allele A_2 produces 5 units of enzyme. Which allele is dominant and which is haploinsufficient? Explain your reasoning. Solution Haploinsufficiency is a potential mechanism in which a diploid species can express a single functional copy of that gene and the other one copy of gene function will be lost due to mutation finally trigger loss-of-function mutation so that it is not possible to generate a complete wild-type phenotype with one functional copy of allele. Therefore, in the above case, enzyme aggase need >16 units of enzyme, in which only A1 generates 12 units & the remaining A2 produced 5 units together 17 units to produce a complete wild-type phenotype. Here, haplosuffciency can occur with a single functional copy of gene, A1 & A2 alleles \"individually\" both are does not produce enough product to meet >16 units of enzyme to display the wild type\'s so that both are considered as haplosufficient Dominance can be observed only in allele A1 because it can mask the allelic recessive locus due to heterozygosity & Mendelian inheritance in which A1 is responsible for the phenotype Other reason: 1. Dominant gain of function in which protein expressed from the mutant gene is constitutively active due to base substitutions finally protein expression is further downregulated because of increase in expression of enzyme proteins. 2. Loss of Function/ Haploinsufficiency: It is also due to loss of heterozygosity thereby by loss of gene function with haploinsufficiency. Finally, resultant organism possesses only one functional gene copy with truncated protein..

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