Question

Consider two unlinked loci, A, a and B, b. The frequencies of the nine genotypes in a population that are recognizable with respect to these loci are:

AA Aa aa

BB 0.20 0.40 0.04

Bb 0.05 0.07 0.20

bb 0.00 0.03 0.01

(a) Apply a procedure to determine whether this population is in equilibrium with respect to each single locus

(b) Apply a procedure to determine if it is in equilibrium with respect to the nine genotypic frequencies.

(a) In equilibrium? Frequency of - A allele = B allele =

(b) In equilibrium?

Answer 1

(a) Apply a procedure to determine whether this population is in equilibrium with respect to each single locus.

In this case we are going to apply a chi square but for each locus independently, but first let's calculate the frequency of each allele:

A frequency: square root of 0.25 = 0.5

a frequency: square root of 0.25 = 0.5

B frequency: square root of 0.64 = 0.8

b frequency: 1square root of 0.04 = 0.2

Now, let's calculate our expected genotype frequences (because we already have the observed ones):

	A/B	a/b
A/B	AA/BB	Aa/Bb
a/b	Aa/Bb	aa/bb

The expected frequencies are:

AA = 0.25

Aa = 0.5

aa = 0.25

BB = 0.25

Bb = 0.5

bb = 0.25

Now, let's apply a chi square for each one, the formula is:

We are going to assume a sample size, because we cannot apply a chi square with frequencies, we need to translate it into occurrances. Let's assume a population of 100, then we will multiply each frequency by 100

Chi for Locus A

	Observed	Expected	(O-E)2/E
AA	25	25	0
Aa	50	50	0
aa	25	25	0
Sum			0

Our sum value is our P value and we would normally compare it to the value from the significance 0.05, but in this case it almost innecesary, since the value is zero it means the population is in equilibrium for Locus A

Chi for Locus B

	Observed	Expected	(O-E)2/E
AA	64	25	60.84
Aa	32	50	6.48
aa	4	25	17.64
Sum			84.96

Now, let's compare our P value with our value in a chi square dsitribution. I'll lend you this one:

Choose one degree of freedom (number of alleles minus 1) and 0.05 significance. The value there is 3.84, our value is larger. That means the population is not in equilibrium for Locus B

(b) Apply a procedure to determine if it is in equilibrium with respect to the nine genotypic frequencies.

Well, now we have to do the same but for the loci together. Let's calculate the genotypic expected frequencies:

	AB	Ab	aB	ab
AB	AABB	AABb	AaBB	AaBb
Ab	AABb	AAbb	AaBb	Aabb
aB	AaBB	AaBb	aaBB	aaBb
ab	AaBb	Aabb	aaBb	aabb

AABB frequency: 1/16 = 0.0625

AABb frequency: 2/16 = 0.125

AAbb frequency: 1/16 = 0.0625

AaBB frequency: 2/16 = 0.125

AaBb frequency: 4/16 = 0.25

Aabb frequency: 2/16 = 0.125

aaBB frequency: 1/16 = 0.0625

aaBb frequency: 2/16 = 0.125

aabb frequency:1/16 = 0.0625

Now, let's apply the chi square, but let's multiply the frequencies by 100 again, due to simulating a population of 100:

Genotype	Observed	Expected	(O-E)2/E
AABB	20	6.25	30.25
AABb	5	12.5	4.5
AAbb	0	6.25	6.25
AaBB	40	12.5	60.5
AaBb	7	25	12.96
Aabb	3	12.5	7.22
aaBB	4	6.25	0.81
aaBb	20	12.5	4.5
aabb	1	6.25	4.41
Sum			131.4

Again we go and check our chi table, our degrees of freedom now are 3 (number of alleles minus 1). So our value to compare now is 7.815. Our value is larger, this means the population is not in equilibrium for the 9 genotypes.