Test Subjects Population Australian Aborigines European Origin African, Asian, and Native American Tasters (p² + 2pq) Number 0.503 Phenotypes Percent Non-tasters (q²) Number 0.497 0.240 0.135 Percent Allele Frequency (based on H-W)
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Test Subjects Population Australian Aborigines European Origin African, Asian, and Native American Tasters (p² + 2pq) Number 0.503
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- I asked this question before but the solution was not correct and I'm really trying to figure out how to do this to get the correct answer. Please show work!DNA typing is used to compare evidence DNA (E) left at a crime scene to two suspects(S1 and S2). Suspect 1 is excluded by the evidence, but suspect 2 remains included. What isthe frequency of suspect 2's genotype if the allelic frequencies in the population are f(A1) =0.1, f(A2) = 0.2, and f(A3) = 0.7, and the population is at Hardy–Weinberg equilibrium?The answer is 0.49.I. Chi-Square Test Phenotype Class Observed Frequency (0) Seed Expected Frequency (E) Deviation d2 d?/E (d=0-E) Corn Tall-normal 26 9 17 289 3.01 (Corn) Rice Tall-wrinkled 26 3 23 529 5.51 (Rice) Mongo Dwarf-normal 25 3 22 256 2.66 (Mongo) Squash Dwarf- 19 3 116 484 5.04 wrinkled (Squash) 16.22 Degree of freedom - 4-1-3 EL X' = S lau -8) 23 G,089 18 %3D X'= 338 Exercise No. 4- Gene Segregation and Interaction 4 What can you conclude based on the value of the computed Chi-square? How can you relate the two principles of Mendel to Chi-Square Values?I need help in table 2 & 3 please Table 1 Phenotype O E O-E (O-E)^2 (O-E)^2 /E Disease, Male 304 433.875 129.875 16789.68 38.69 Disease, Female 267 433.875 166.875 26855.01 61.89 WT, Male 285 144.625 140.375 19705.14 136.25 WT, Female 301 144.625 156.375 24453.15 156.37 Total 1157 1157 393.20 DF 3 p-value 7.84 Expected progencies as per SLR MOI’s = 1:1 for both male and female Phenotype O E O-E (O-E)^2 (O-E)^2 /E Disease, Male 304 289.25 14.75 217.56 0.75 Disease, Female 267 289.25 22.25 495.06 1.85 WT, Male 285 289.25 4.25 18.06 0.06 WT, Female 301 289.25 11.75 138.06 0.46 Total 1157 1157 3.12 DF 3 p-value 7.84 Progenies are following SR MOI’s that parents have genotypes and progenies in ratio 1:1 In Table 2,…
- Once upon a time there was a population of hippies living in an isolated commune called Gone. They had been there for generations, and all members were homozygous for a gene that gave them a preference for crafting necklaces out of green beads, so that every individual had the genotype GG at this locus. In another isolated commune, called Yonder, was a population of hippies that had a genetic propensity for making necklaces out of yellow beads, and everyone was homozygous for a different allele at the same genetic locus; their genotype was YY. Last year (prior to our current social isolating) everyone repaired their VW buses, and a giant gathering in celebration of the 50th anniversary was held at the Oregon Country Fair. The next few questions will ask you about these two populations that came together at the fair.The Distribution and Chi Square Test of Sexes in Families of Four Children D/E Observed number Expected number of families (0) Combination Deviation (d) of families All boys (b 3 boys & 1 girl (b3 g) 2 boys & 2 girls (b'g) 1 boys & 3 girls (bg³) All girls (g*) ТОTAL: Computed chi square value: Tabular chi square value : df and at 0.05 level of significance DecisionCase 7514997 CHILD Name Alleged FATHER Test No. 7514997-20 7514997-30 Locus PI Allele Sizes Allele Sizes 03S1358 DIS1656 D2S441 D1OS1248 0.83 D135317 Penta E 016S539 D18S51 D2S1338 CSF 1PO Penta D 0.00 16 16 10 13 17 17 14 14 18 16.3 14 14 0.00 18.3 1.93 16 12 16 12 0.00 11 10 11 10 12 17 10 10 0.00 10 15 13 0.00 11 23 0.00 14 0.00 25 11 13 14 6. 0.00 1.12 12 THOI 1.04 17 31.2 wWA 0.00 14 32.2 021S11 D7S820 DSS8 18 TPOX D8S1 179 0128301 D198433 FGA 022$ 1045 0.00 Antogenin 0.00 0.00 11 11 12 11 0.67 1.86 11 14 0.00 16 0.00 0.00 14 19 11 4.13 19 15 12 Determine if this is the father and explain your conclusion. 100%
- - 1 attacnment Phenotype Genotype Number MM LMLM 182 MN LMLN 172 NN LNLN 44 Calculate the genotypic frequency of the LMLN genotype? (2 pts) A. 0.457 C. 0.111 B. 0.432 D. 0.216Trivla Game Show _Make Your Own Tri ngston.schoology.com/common-assessment-delivery/start/4789189591?action=onresume&submissionld=463322566 Dillon WF g Aa v Done In guinea pigs, black hair (B) is dominant to white hair (b) and rough hair (R) is dominant to smooth hair (r). What are all the possible genotypes of a guinea pig that has black, rough hair? (Select all that apply.) O BBRR BBRr BBrr BBRR BbRr O bbRR O bbRr O bbrr O Black O White O Rough OSmooth O Rough O SmoothStyles Dictate Sensitivity Pane Problem #2 In sheep, white wool (W) is dominant over black wool (w). What genotype and phenotype ratios are expected from a cross between a heterozygous sheep with white wool and a sheep with black wool? P Genotypic Ratio: Phenotypic Ratio: Editor
- A BOOKMARK Question 21/28 E. In reference to chart 1, what is the probability or chance that the offspring of the two pea pods will be Use the diagrams below to answer the questions that follow. 21 Pea Plants green in color? Chart 1 A) 25% Punnett Square %0 c) 75% Aa A a %00 Aa Aa %0 (3 Chart 2 AaBb x AaBb AB aB ab Key AB AABB AABB Ab AABB AAbb poous - 88 AA - Green AABB AaBb AaBb Aabb aB AaBB AaBb ab AaBb Aabb aa - Yellow aaBB aaBb aaBb aabb 46noy - 99How to caculate the frequency for genotype and phenotype Student # Genotype (4/4,⅘,5/5) Chronotype number Phenotype (Morning 59-86, Intermediate 42-58,Evening 16-41) Student 1 4,5 51 Intermediate Student 2 5,5 33 Evening Student 3 4,5 40 Evening Student 4 5,5 52 Intermediate Student 5 5,5 51 Intermediate Student 6 4,4 39 Evening Student 7 4,4 46 Intermediate Student 8 4,4 48 Intermediate Student 9 4,5 59 Morning Student 10 5,5 31 Evening Student 11 4,5 39 Evening Genotype # of students Frequency Phenotype # of students Frequency 4/4 3 ? Morning 59-86 1 ? 4/5 4 Intermediate 42-58 5 5/5 4 Evening 16-41 5The following data were obtained from a Ghanaian population of 152 people. Blood Type Number of Individuals M 61 MN 64 N 27 Calculate the frequencies of the M and N alleles in this population. What are the expected numbers of individuals of each genotypic class in this population? Assume Hardy-Weinberg conditions. Use the chi-squared test to determine if these data fit the Hardy-Weinberg equilibrium model. The degrees of freedom for this test should be 1. Why is this appropriate?