Question 14 cl 16 Sapling Learning mental retardation if by a allele and can cau

Question

Question 14 cl 16 Sapling Learning mental retardation if by a allele and can cause a severe form of United detected The prevalence of PKU aa) in the States according to the National Institutes of Health (200 Given the above data, what is the expected alele frequency ofthe a allele? Number freq(a) 0.00816 Suppose the following data was collected for newborn PKU incidences in a separate population, the state of Vermont, in a single year. Number of Genotype incidences PKU negative 40,422 PKU negative 8,000 PKU positive aa 700 49122 Total: Determine if the population of Vermont is in Hardy-Weinberg equilibrium, by calculating the chi-square (x) value given the above data. Number Continued below. Previous 8 Give Up & View solution check Answer 0 Next Exi Tools Hint

Explanation / Answer

a) 1 out of 15,000 newborns has PKU. This newborn will have genotype 'aa'.

q^2 = 1/15000; q= allele frequency of 'a' = 0.0007

b) Suppose there is a cross between 'Aa' and 'Aa'

Fitness of 'aa' is least. So, it is totally clear that HW equilibrium is not followed.

Using Chi-square,

Null hypothesis: Ho = the population is in HW equilibrium

Observed

Expected

(O-E)^2/E= chi square value

AA

40422

12280.5

64487.93

Aa

8000

24561

11166.75

aa

700

12280.5

10920.4

Total

49122

86575.08= calculated value

Degree of freedom = (rows-1) (columns-1) = (3-1)* (3-1) = 4

Tabulated value = taken from chi square table = 9.49 (Assuming 0.05 level of significance)

Null hypothesis is accepted if calculated value is less than tabulated value.

Here, tabulated value<<<<calculated value, so, null hypothesis should be rejected.

So, the population is not in HW equilibrium.

Observed

Expected

(O-E)^2/E= chi square value

AA

40422

12280.5

64487.93

Aa

8000

24561

11166.75

aa

700

12280.5

10920.4

Total

49122

86575.08= calculated value

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