For Sample 1 conduct a chisquare test of the hypothesis that

For Sample 1, conduct a chi-square test of the hypothesis that the pathogen population is in Hardy-Weinberg equilibrium. What is the chi-square test value for this sample?

3.84

0

0.25

0.5

9.07

For Sample 2, conduct a chi-square test of the hypothesis that the pathogen population is in Hardy-Weinberg equilibrium. What is the chi-square test value for this sample?

0.5

0

9.07

3.84

0.25

	a.	3.84
	b.	0
	c.	0.25
	d.	0.5
	e.	9.07

A human pathologist obtains multiple cultures of a diploid pathogen taken from two different parts of the body of a single infected human individual (samples 1 and 2). Using gel electrophoresis, each culture is screened for its genotype at a locus affecting its virulence (i.e., nastiness) to the host. The cultures are found to be segregating for fast (F and slow migrating (S) alleles. Here are the total counts of genotypes the researcher found in each sample: Genotypes FF SS 420 Sample 1 90 490 110 380 Sample 2 510

Solution

When there are two alleles for a particular gene--F and S and their respective population frequencies are p and q, then the expected frequencies of the genotypes FF, FS, and SS
FF = p²,     FS = 2pq,    SS = q²

For sample 1

F allele frequency = ((2 * 90) + 420) / (2 * 1000) = 0.3

S allele frequency = ((2 * 490) + 420) / (2 * 1000) = 0.7

If individuals have no genotype preference when mating and there is no significant level of allele mutation, then the expected frequencies of the genotypes FF, FS, and SS are

FF: (0.3)(0.3) = 0.09

FS: 2(0.7)(0.3) = 0.42
SS: (0.7)(0.7) = 0.49

We perform the chi-square test to the observed and expected allele values

Observed

Expected

FF

90

90

FS

420

420

SS

490

490

Chi squre value = (90 -90 )² / 90 + (420 -420) ²/420 + (490-490)²/490 = 0

Therefore, ² = 0 < 3.841 we will accept the null model and conclude that there is significant statistical support that the population is in Hardy-Weinberg equilibrium.

For sample 2

F allele frequency = ((2 * 110) + 380) / (2 * 1000) = 0.3

S allele frequency = ((2 * 510) + 380) / (2 * 1000) = 0.7

If individuals have no genotype preference when mating and there is no significant level of allele mutation, then the expected frequencies of the genotypes FF, FS, and SS are

FF: (0.3)(0.3) = 0.09

FS: 2(0.7)(0.3) = 0.42

SS: (0.7)(0.7) = 0.49

We perform the chi-square test to the observed and expected allele values

Observed

Expected

FF

110

90

FS

380

420

SS

510

490

Chi squre value = (110 -90 )² / 90 + (380 -420) ²/420 + (510-490)²/490 = 9.07

Therefore, ² = 9.07 > 3.841 we will reject the null model and conclude that there is significant statistical support that the population is not in Hardy-Weinberg equilibrium.

	Observed	Expected
FF	90	90
FS	420	420
SS	490	490