Which of the following would be sufficient for the Hardy-Weinberg equ.pdf

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Which of the following would be sufficient for the Hardy-Weinberg equation to accurately predict genotype frequencies from allele frequencies? p + q = 1 the population is not evolving due to natural selection. the population is not evolving due to any of the conditions that disrupt Hardy-Weinberg equilibrium. the population is infinitely large. in a village, if the proportion of individuals who have sickle-cell disease is 0.10, and the population is assumed to be at Hardy- Weinberg equilibrium, what is the expected frequency of the HbS allele? 0.32 0.18 0.90 0.01 in a village where the proportion of individuals who are susceptible to malaria (genotype HbA/HbA) is 0.31, and the population is assumed to be at Hardy-Weinberg equilibrium, what proportion of the population should be heterozygous HbA/HbS? Solution Ans. 5.5. Correct option- C Option A. Incorrect. A population NOT at Hardy-Weinberg (HW) equilibrium also has p + q= 1. Since the frequencies p, and q are calculated considering their relative abundance, their sum of always 1.0. However, the value of p and q changes over generations when population is evolving. The value of p and q remains constant over generations if the population is at HW equilibrium. Option B. Incorrect. Factors other than natural selection, like genetic drift, bottle-neck effect, etc. also disrupt HW equilibrium. So, even if natural selection does not occur, the population may evolve through genetic drift, etc. Option C. Correct. HW equation holds true when no factor causes evolution of the population. Option D. Incorrect. HW equation holds true when no factor causes evolution of the population. Being a large population do not exclude the possibilities of evolution. Ans. 5.6. Normal, dominant allele = HbA Mutant, recessive allele = Hbs Normal, homozygous genotype = HbA HbA - susceptible to malaria Normal, heterozygous genotype = HbA Hbs - it is a carrier, but unaffected by disease Recessive, affected genotype = Hbs Hbs - affected with sickle-cell disease Now, Hardy- Weinberg Equation for 2 allele system (for a population at HW equilibrium) p + q = 1 - equation 1 (p + q)2 = p2 + q2 + 2pq = 1 - equation 2 Where, p = allelic frequency of dominant allele HbA q = allelic frequency of recessive allele HbS p2 = genotypic frequency of HbA HbA q2 = genotypic frequency of Hbs Hbs 2pq = genotypic frequency of HbA Hbs Given, Frequency of Hbs Hbsindividuals = q2 = 0.10 Or, q2 = 0.10 Or, q = (0.10)1/2 = 0.316 = 0.32 ; [Note: (0.10)1/2 is under root (.010)] Hence, frequency of allele HbS = q = 0.32 Correct option. A. Ans. 5.7 Given, Frequency of HbA HbAindividuals = p2 = 0.31 Or, p = (0.31)1/2 = 0.55677 Hence, frequency of HbA allele = 0.55677 Putting the value of p in equation 1- p + q = 1 or, q = 1 – 0.55677 = 0.44323 Hence, q = 0.44323 Now, Frequency of the heterozygotes (HbA Hbs) = 2 pq = 2 x (0.55677) x (0.44323) = 0.493 Hence, frequency of heterozygotes = 0.493 % of heterozygotes = 49.3 %.

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q = allelic frequency of recessive allele HbS
p2 = genotypic frequency of HbA HbA
q2 = genotypic frequency of Hbs Hbs
2pq = genotypic frequency of HbA Hbs
Given,
Frequency of Hbs Hbsindividuals = q2 = 0.10
Or, q2 = 0.10
Or, q = (0.10)1/2 = 0.316 = 0.32 ; [Note: (0.10)1/2 is under root (.010)]
Hence, frequency of allele HbS = q = 0.32
Correct option. A.
Ans. 5.7
Given,
Frequency of HbA HbAindividuals = p2 = 0.31
Or, p = (0.31)1/2 = 0.55677
Hence, frequency of HbA allele = 0.55677
Putting the value of p in equation 1-
p + q = 1
or, q = 1 – 0.55677 = 0.44323
Hence, q = 0.44323
Now,
Frequency of the heterozygotes (HbA Hbs) = 2 pq = 2 x (0.55677) x (0.44323) = 0.493
Hence, frequency of heterozygotes = 0.493
% of heterozygotes = 49.3 %

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Question: In C Programming: In mathematics, a set is a colle... Save In C Programming: In mathematics, a set is a collection of distinct elements (say integer numbers). For example, A={2, 5, 7} and B={3, 5, 8, 10} are two different sets. There are several basic operations for constructing new sets from given two sets, but let\'s just consider three basic ones: C = union(A,B); ---> C = A B={2, 5, 7, 8, 10} C = intersection(A,B); ---> C = A B={5} C = difference(A,B); ---> C = A-B = A \\ B ={2, 7} also aka. complement You are asked to develop a set library (using two different representations: array and link list) and then implement a driver program that gets two sets and one of the above operation as a command to apply, then it prints the resulting set... Developing set library: For interface, create an interface file set.h which contains boiler plate and the followings: typedef int setElementT; typedef struct setCDT *setADT; setADT setNew(); /* create a new empty set */ void setFree(setADT S); /* free the space allocated for the set S */ int setInsertElementSorted(setADT S, setElementT E); /* if not successful, return 0; otherwise, return the num of elements after the insertion. Also note that the elements might be given in different orders, but your function should always keep the set in a sorted manner after each insertion */ setADT setUnion(setADT A, setADT B); /* returns a new set containing A B */ setADT setIntersection(setADT A, setADT B); /* returns a new set containing A B */ setADT setDifference(setADT A, setADT B); /* returns a new set containing A \\ B */ int setCardinality(setADT S); /* return the number of elements in S */ void setPrint(setADT S, char *name); /* print elements of S, A = {2, 5, 7} */ For implementation, you will be asked to have two different implementations: (40pt) First implement set library as setArrayImp.c which uses a constant size array to store set elements (suppose max set size is 100). [hint: see ch2 slides 72-78 ] (40pt) Second implement this library as setLinkedListImp.c which uses a dynamic single linked list to store set elements. Develop a driver.c program and compile it with two different imp of set library (20 pt) 1. Create two sets called A and B. 2. Ask user to enter positive integers for set A (end input when user enters -1) 3. Ask user to enter positive integers for set B (end input when user enters -1) 4. In a loop 4.1. Ask user to enter a command: 4.2 If Q is entered, quit from this loop. 4.3 If U, I, or D is entered, compute set C as union, intersection, or difference. setPrint(A, \"A\"); setPrint(A, \"B\"); setPrint(C, \"C\"); print the number of elements in C setFree(C); 5. free A and B Compile/execute driver.c with setArrayImp.c as well as setListImp.c //no include files needed Solution Create set.h first with the following typedef int setElementT; typedef struct setCDT *setADT; setADT setNew(); /* Creates a new empty set */ void setFree(setADT S); /* Frees the space allocated for the set .

Question In C Programming In mathematics, a set is a colle...Sav.pdf

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Q1. What is the primary reason that Fender\'s blue butterflies are endangered? The recent increase in fire frequency and severity has greatly reduced habitat quality. The butterfly\'s primary host plant, Kincaid\'s lupine, became extinct in the 1930s. Increases in disease prevalence and predation have diminished their numbers. Urbanization and agriculture have reduced and fragmented the butterfly\'s preferred habitat. Solution Option (d) is correct: Urbanization and agriculture have reduced and fragmented the butterfly\'s prefered habitat. Fender\'s blue butterfly is an endangered subspecies of butterfly endemic to the Willamette Valley of northwestern Oregon, United States. The prime reason for its population decline is that its native prairie habitat has been largely converted to agriculture, subjected to fire suppression and invaded by non-native plants. With massive changes in the fire regime, disturbances that maintained native prairies have been greatly altered, allowing tree and shrub species to invade and shade out low-growing Kincaid\'s lupine, the species upon which this butterfly depends. Also,the non-native species such as Himalayan blackberry has aggressively overtaken open spaces. Loss of native prairie habitat has resulted in the isolation of butterfly populations. Thus, these butterfly populations are more vulnerable to natural and human-made disturbances. As the number of sites have declined and the distance between them has increased and the opportunities for adult movement between populations are reduced, which were earlier inter-connected. Thus the Fender\'s blue butterflies populations is isolated, has become endangered and faces a higher risk of extinction..

Q1. What is the primary reason that Fenders blue butterflies are e.pdf

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Normality assumption is very important. A researcher is interested in studying body temperature. He collected the data below from 10 persons. Could you please as a statistician check the data follow Normal distribution or not using Normal quantile plot? 96.3 96.7 96.9 97.0 97.1 97.1 97.2 97.4 95.5 97.1 Solution The normal probability plot is The points are in the form of a straight line. therefore the given data is normal..

Normality assumption is very important. A researcher is interested i.pdf

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Punnett square: Father and mother are heterozygous and each are carriers for a dominant linked disorder. What is the % chance the child will be born with the disorder Punnett square: Father and mother are heterozygous and each are carriers for a dominant linked disorder. What is the % chance the child will be born with the disorder Punnett square: Father and mother are heterozygous and each are carriers for a dominant linked disorder. What is the % chance the child will be born with the disorder Solution parents A a A AA Aa a Aa aa Since parents are heterozygotic carriers their phenotype is Aa Aa The gamets are A a Progeny 25%AA 50%Aa 25%aa ie 1:2:1 25% chance for having the disease(AA) parents A a A AA Aa a Aa aa.

Punnett square Father and mother are heterozygous and each are .pdf

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Property Values Is this rooted tree Leaf vertices Intenal vertices A.pdf

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