Question

I need serious help with the chi-square part of this problem (X²). I have posted this problem before and gotten the incorrect answer, looked up previously answered questions that were similar, and tried it myself more than a few times. If someone can please help and show a fairly detailed step process as to how to figure this out, that would be greatly appreciated.

Mapt Phenylketonuria (PKU) is caused by homozygosity for a recessive allele and can cause a severe form of mentál retardation if not detected early. The prevalence of PKU (genotype aa) is 1 per 15,000 newborns in the United States according to the National Institutes of Health (2001) Given the above data, what is the expected allele frequency of the a allele? Number freq(a)= 10.008 16 Suppose the following data was collected for newborn PKU incidences in a separate population, the state of Vermont, in a single year Number of 20,212 4,000 350 24562 Phenotype PKU negative PKU negative PKU positive Aa Total Determine if the population of Vermont is in Hardy-Weinberg equilibrium, by calculating the chi-square (X) value given the above data Incorrect. To calculate the chi-square value, you will need to first calculate the expected number of incidences by determining the frequencies of the a and A alleles. Next, use the chi-square equation to determine the chi-square value for each of the genotypes. Remember to add together the Number 27304.39 Continued below...

Answer 1

Answer:

1).

Frequency of PKU = q^2 = 1/15000 = 0.000067

Frequency of PKU allele = q = SQRT of q^2 = 0.0081

2).

Total number of A alleles = 2 *AA + Aa

= 2 * 20212 + 4000 = 44424

Total number of a alleles = 2 * aa + Aa

= 2 * 350 + 4000 = 4700

Total alleles = 49124

Frequency of allele ‘A’ = 44424 / 49124 = 0.904

Frequency of allele ‘a’ = 4700 / 49124 = 0.096

Expected AA individuals = 0.904 * 0.904 * 24562 = 20073

Aa individuals = 2 * 0.904 * 0.096 * 24562= 4263

aa individuals = 0.096 * 0.096 * 24562 = 226

Phenotype	Observed(O)	Expected (E)	O-E	(O-E)2	(O-E)2/E
AA	20212	20073	139	19321.00	0.9625
Aa	4000	4263	-263	69169.00	16.2254
aa	350	226	124	15376.00	68.0354
	24562	24562			85.2234

Chi-square value = 85.22

Degrees of freedom = number of phenotypes – 1

Df = 3-1=2

Chi-square value of 85.22 is greater than the critical value of 5.99. We can reject the null hypothesis and the population is not in Hardy-Weiberg equilibrium.