Table 3. Genotypic Frequencies for Alu in a USA Sample. Category Number Frequecy
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Table 3. Genotypic Frequencies for Alu in a USA Sample. Category Number Frequecy Homozygous(+/+) 2,4222 0.24 Heterozygous(+/-) 5,528 0.55 Homozygous(+/+) 2,050 0.21 total=10,000 =1.00 Table 4. Allelic Frequencies for USA Category Number Frequency (+) alleles 10372 p= 0.5186 (-) alleles 9628 q= 0.4814 20,000 =1.00 Using the values for p and q that you calculated in Table 4for the USA population sample, calculate p2, 2pq, andq2. Do they come out to be the same as the genotypefrequencies that you found in Table 3? Does this USA-wide sample suggest that the population of theUSA is in Hardy-Weinberg equilibrium? Table 3. Genotypic Frequencies for Alu in a USA Sample. Category Number Frequecy Homozygous(+/+) 2,4222 0.24 Heterozygous(+/-) 5,528 0.55 Homozygous(+/+) 2,050 0.21 total=10,000 =1.00 Table 4. Allelic Frequencies for USA Category Number Frequency (+) alleles 10372 p= 0.5186 (-) alleles 9628 q= 0.4814 20,000 =1.00 Using the values for p and q that you calculated in Table 4for the USA population sample, calculate p2, 2pq, andq2. Do they come out to be the same as the genotypefrequencies that you found in Table 3? Does this USA-wide sample suggest that the population of theUSA is in Hardy-Weinberg equilibrium? Category Number Frequecy Homozygous(+/+) 2,4222 0.24 Heterozygous(+/-) 5,528 0.55 Homozygous(+/+) 2,050 0.21Explanation / Answer
From the first table we know that p2 = 0.24
q2 = 0.21
2pq = 0.55
from the second table we see that p = 0.5186
which gives p2 =0.2689
q = 0.4814 which gives q2 = 0.2317
2pq = 0.4993
by using chi-square test ,difference between two samples can becompared.
As there is no difference betweenabove two. , this population is in Hardy-WeinbergEquilibrium
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