DATA SET 01The number of offspring in F1 and F2 generations: 510The true mode of inheritance is autosomal dominant.Parental cross: Father with disease phenotype, Mother with wild-type phenotype. F1 COUNTS FOR DATA SET 01PhenotypeGender. Disease Wild-Type Male 237 0Female 273 0 F2 COUNTS FOR DATA SET 01PhenotypeGender Disease Wild-Type Male 185 73Female 183 69 DATA SET 02The number of offspring in F1 and F2 generations: 899F1 COUNTS FOR DATA SET 02PhenotypeGender Disease Wild-Type Male 424 0Female 475 0 F2 COUNTS FOR DATA SET 02PhenotypeGender Disease Wild-Type Male 349 112Female 323 115F2 COUNTS FOR DATA SET 01GenderPhenotypeObserved counts (O) Expected Proportions Expected Counts (E) (O-E) (O-E)^2/E MaleWT73 MaleDisease185 FemaleWT69 FemaleDisease183 Total 510 DegFrdm P-value F2 COUNTS FOR DATA SET 02GenderPhenotypeObserved counts (O) Expected Proportions Expected Counts (E) (O-E) (O-E)^2/E MaleWT112 MaleDisease349 FemaleWT115 FemaleDisease323 Total 899 DegFrdm P-value Can you please do the same in this example and fill in those two tables, please typed The solution should be like this one ???????? The number of offspring in F1 and F2 generations: 854Punnett Square for Fi parents.Mother's allelesFather's allelesd.dDDdDddddddFrom intro, we know that WT must be homozygous, and because we are dealing with AutosomalDominant, Homozygous Lethal, other parent must be Dd. This table is identical to parental Punnettsquare.So, according to Punnett Square, F1 generation has 50% disease, 50% WT. Multiply by ½ to get male andfemale, or 25% for each.F1 COUNTS FOR DATA SET 02PhenotypeGender Disease Wild-TypeMale 209213Female 223209F2 COUNTS FOR DATA SET 022 |PagePhenotypeGender Disease Wild-TypeMale 219182Female 213240F2 CHI-SQUARE TEST ANALYSIS WITH TRUE MOITable for compute X^2 Goodness of Fit valuePhenotypeObsCounts(O) ExpPropExpCounts(E) (О-Е)213,5213,5213,5213,5(O-E)^2/E-31.55.526,5MaleWT1822194.64750.141690.25Male Disease0.25FemaleFemale DiseaseWT2400.253.28922130.25-0.50.001171X^2 Goodness-of-fit Statistic : 8.07962529Degrees of freedomP-value:3:0.0443944375P-value < 0.05. Reject null hypothesisfor your specified mode of inheritance.So, data set most consistent with MOI is Data Set 01, since we do not reject the null hypothesis, and wedo in Data Set 02. Parental cross: Father with disease phenotype, Mother with wild-type phenotype.F1 COUNTS FOR DATA SET 01Punnett Square for Parents:Mother's allelesFather's allelesddDDdDddddddAccording to Punnett Square, F1 generation has 50% disease, 50% WT. Multiply by ½ to get male andfemale.PhenotypeGender Disease Wild-TypeMale 133133Female 118118Punnett Square for Parents:Mother's allelesFather's allelesdDdDddPPddddFrom intro, we know that WT must be homozygous, and because we are dealing with AutosomalDominant, Homozygous Lethal, other parent must be Dd. This table is identical to parental Punnettsquare.So, according to Punnett Square, F1 generation has 50% disease, 50% WT. Multiply by ½ to get male andfemale, or 25% for each.1| PageF2 COUNTS FOR DATA SET 01PhenotypeGender Disease Wild-TypeMale 123121Female 144114F2 CHI-SQUARE TEST ANALYSIS WITH TRUE MỌITable for compute X^2 Goodness of Fit valuePhenotypeObsCounts(O) ExpPropExpCounts(E) (O-E)125.5(O-E)^2/E0.161350.049801MaleWT1210.25-4.5-2.5-11.5Male Disease1230.25125.5Female WT1140.25125.51.0538Female Disease1440.25125.518.52.7271X^2 Goodness-of-fit Statistic: 3.99203187Degrees of freedomP-value: 30.26232583P-value > 0.05. Do not reject null hypothesis for your specified mode of inheritance.DATA SET 02The number of offspring in F1 and F2 generations: 854

Name: DATA SET 01The number of offspring in F1 and F2 generations: 510The tr
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Description: DATA SET 01The number of offspring in F1 and F2 generations: 510The true mode of inheritance is autosomal dominant.Parental cross: Father with disease phenotype, Mother with wild-type phenotype. F1 COUNTS FOR DATA SET 01PhenotypeGender. Disease

The default assumptions created by researchers which then is either rejected or accepted by…

DATA SET 01 The number of offspring in F1 and F2 generations: 510 The true mode of inheritance is autosomal dominant. Parental cross: Father with disease phenotype, Mother with wild-type phenotype. F1 COUNTS FOR DATA SET 01 Phenotype Gender. Disease Wild-Type Male 237 0 Female 273 0 F2 COUNTS FOR DATA SET 01 Phenotype Gender Disease Wild-Type Male 185 73 Female 183 69 DATA SET 02 The number of offspring in F1 and F2 generations: 899 F1 COUNTS FOR DATA SET 02 Phenotype Gender Disease Wild-Type Male 424 0 Female 475 0 F2 COUNTS FOR DATA SET 02 Phenotype Gender Disease Wild-Type Male 349 112 Female 323 115 F2 COUNTS FOR DATA SET 01 Gender Phenotype Observed counts (O) Expected Proportions Expected Counts (E) (O-E) (O-E)^2/E Male WT 73 Male Disease 185 Female WT 69 Female Disease 183 Total 510 DegFrdm P-value F2 COUNTS FOR DATA SET 02 Gender Phenotype Observed counts (O) Expected Proportions Expected Counts (E) (O-E) (O-E)^2/E Male WT 112 Male Disease 349 Female WT 115 Female Disease 323 Total 899 DegFrdm P-value Can you please do the same in this example and fill in those two tables, please typed The solution should be like this one ????????

Human Anatomy & Physiology (11th Edition)

11th Edition

ISBN:9780134580999

Author:Elaine N. Marieb, Katja N. Hoehn

Publisher:Elaine N. Marieb, Katja N. Hoehn

Chapter1: The Human Body: An Orientation

Section: Chapter Questions

Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...

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Gender	Phenotype	Observed counts (O)	Expected Proportions	Expected Counts (E)	(O-E)	(O-E)^2/E
Male	WT	73
Male	Disease	185
Female	WT	69
Female	Disease	183
Total		510
					DegFrdm
					P-value