Question

Question 2: A blood sample was drawn from individuals in Finland and the following genotype frequencies were observed among the people sampled: FF-99, Ff-418 and ff-483. The critical value to reject is 3.814. (a) Is this population in Hardy-Weinberg equilibrium? (b) Which statistical test did you perform? SHOW ALL WORK- formulae and calculations.

Answer 1

Given:

Total population: 99+418+483 =1000

Number with genotype FF = 99

Number with genotype Ff = 418

Number with genotype ff = 483

Observed frequencies:

Genotype frequency of FF: 99/1000 = 0.099

Genotype frequency of Ff: 418/1000 = 0.418

Genotype frequency of ff: 483/1000 = 0.483

Expected allele frequencies:

Frequency of F allele =[99 + ½ (418)]/1000 = 0.308 = 0.3 (approx)

Frequency of f allele =[483+ ½ (418)]/1000= 0.692 = 0.7(approx.) =

Expected genotype frequencies:

· Genotype frequency of FF: p2= 0.09

· Genotype frequency of Aa: 2pq = 0.42

· Genotype frequency of aa: q2 = 0.49

Expected numbers:

· Number with genotype FF = 0.09 x 1000 =90

· Number with genotype Ff = 0.42 x 1000 = 420

· Number with genotype ff = 0.49 x 1000 = 490

	O	E	O-E =d	d2	Χ2 =(d2/E)
FF	99	90	9	82	0.911
Ff	418	420	-2	4	0.0095
ff	482	490	8	64	0.130
Total	1000	1000			1.05

· Critical value for df= 1, p= 3.814

· Χ2value in this case= 1.05

·a. Thus, the population is under H-W equilibrium

b.

In Chi square (X2 ) fitness test:

· A hypothesis is considered.

· In an experiment:

· Observed frequency or result (O)- is compared to- expected frequency or result (E).

· The difference or deviation is noted as d.

· The hypothesis is then considered valid or rejected.

· (O-E)2/ E = d2/E

· (O- E)2 = d2

· Degrees of freedom (df) – number of classes/ outcome/variables considered -1 = n-1.

· Significance level represent statistical analysis of an experimental result variation to have occurred by chance or due to errors.