Observed Genotypic Frequencies for theclass. Total number of people=15 Homozygou
ID: 5248 • Letter: O
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Observed Genotypic Frequencies for theclass. Total number of people=15 Homozygous(+/+)= 2 , 0.133 of genotypicFrequencies Heterozygous(+/-)=5 , 0.333 ofgenotypic Frequencies Homozygous (-/-)=8 , 0.5333 ofgenotypic Frequencies (+) alleles =9, P= 0.3 (Allelic Frequencies for the class) (-) alleles=21, q= 0.7 (AllelicFrequencies for the class) Genotypic Frequencies for Alu in a USA sample(USA-wide radom population study) total number of people= 10,000 Homozygous(+/+)= 2,422 , 0.24 of genotypicFrequencies for Alu in USA Heterozygous(+/-)= 5,528 , 0.55of genotypicFrequencies for Alu in USA Homozygous (-/-)=2,050 , 0.21 of genotypic Frequenciesfor Alu in USA (+) alleles =10372, P=0.5186 (Allelic Frequencies for USA) (-) alleles=9628, q= 0.4814 (Allelic Frequencies for USA) Question. How do your actual class data for genotypic and allelicfrequencies compare with those of the random sampling of the USApopulation?Would you expecte them to match? What reasons can youthink of to explain the differences or similarities? Observed Genotypic Frequencies for theclass. Total number of people=15 Homozygous(+/+)= 2 , 0.133 of genotypicFrequencies Heterozygous(+/-)=5 , 0.333 ofgenotypic Frequencies Homozygous (-/-)=8 , 0.5333 ofgenotypic Frequencies (+) alleles =9, P= 0.3 (Allelic Frequencies for the class) (-) alleles=21, q= 0.7 (AllelicFrequencies for the class) Genotypic Frequencies for Alu in a USA sample(USA-wide radom population study) total number of people= 10,000 Homozygous(+/+)= 2,422 , 0.24 of genotypicFrequencies for Alu in USA Heterozygous(+/-)= 5,528 , 0.55of genotypicFrequencies for Alu in USA Homozygous (-/-)=2,050 , 0.21 of genotypic Frequenciesfor Alu in USA (+) alleles =10372, P=0.5186 (Allelic Frequencies for USA) (-) alleles=9628, q= 0.4814 (Allelic Frequencies for USA) Question. How do your actual class data for genotypic and allelicfrequencies compare with those of the random sampling of the USApopulation?Would you expecte them to match? What reasons can youthink of to explain the differences or similarities? total number of people= 10,000 Homozygous(+/+)= 2,422 , 0.24 of genotypicFrequencies for Alu in USA Heterozygous(+/-)= 5,528 , 0.55of genotypicFrequencies for Alu in USA Homozygous (-/-)=2,050 , 0.21 of genotypic Frequenciesfor Alu in USA (+) alleles =10372, P=0.5186 (Allelic Frequencies for USA) (-) alleles=9628, q= 0.4814 (Allelic Frequencies for USA) Question. How do your actual class data for genotypic and allelicfrequencies compare with those of the random sampling of the USApopulation?Would you expecte them to match? What reasons can youthink of to explain the differences or similarities? total number of people= 10,000 Homozygous(+/+)= 2,422 , 0.24 of genotypicFrequencies for Alu in USA Heterozygous(+/-)= 5,528 , 0.55of genotypicFrequencies for Alu in USA Homozygous (-/-)=2,050 , 0.21 of genotypic Frequenciesfor Alu in USA (+) alleles =10372, P=0.5186 (Allelic Frequencies for USA) (-) alleles=9628, q= 0.4814 (Allelic Frequencies for USA) Question. How do your actual class data for genotypic and allelicfrequencies compare with those of the random sampling of the USApopulation?Would you expecte them to match? What reasons can youthink of to explain the differences or similarities? How do your actual class data for genotypic and allelicfrequencies compare with those of the random sampling of the USApopulation?Would you expecte them to match? What reasons can youthink of to explain the differences or similarities?Explanation / Answer
you are using gene count method orsquare root method for calculation ofgenotypic and allelic frequencies compare with those of the randomsampling of the USA population. Hardy weinberg principle ..........allcondition not there in USA POPULATION,like inbreeding ,initialfrequencies...etc... that is why there is difference between observed and expectedfrequencies in the population
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