Question

Testing the goodness of fit between the data and the Hardy Weinberg equilibrium model generated expectations.

4.1 In a species of bird, feather color is controlled by genes at a single locus, with the red feather allele dominant to the yellow feather allele. A population has 22 red and 14 yellow birds, with 9 of the red birds having a homozygous dominant genotype. Is this population in equilibrium?

Calculate p and q from the number of individuals of each genotype:

p = ____

q = ____

Calculate the expected frequency of each genotype if the population is in equilibrium:

_____ = Frequency of homozygous dominant individuals

_____ = Frequency of heterozygous individuals

_____ = Frequency of homozygous recessive individuals

Calculate the expected number of individuals of each genotype in a population of 36 birds if the gene is in equilibrium:

_____   = Number of homozygous dominant individuals

_____   = Number of heterozygous individuals

_____ = Number of homozygous recessive individuals

Test how well your data fits the expected values from the equilibrium model:

_____ = Chi-square test statistic

_____ = P value

_____ (y/n) in equilibrium?

4.2 In a species of mouse, tail length is controlled by genes at a single locus, with the long tail allele dominant to the short tail allele. A population has 49 long tail and 25 short tail mice, with 22 of the long tail mice having a homozygous dominant genotype. Is this population in equilibrium?

Calculate p and q from the number of individuals of each genotype:

p = _____

q = _____

Calculate the expected frequency of each genotype if the population is in equilibrium:

_____   = Frequency of homozygous dominant individuals

_____   = Frequency of heterozygous individuals

_____   = Frequency of homozygous recessive individuals

Calculate the expected number of individuals of each genotype in a population of 74 mice if the gene is in equilibrium:

_____   = Number of homozygous dominant individuals

_____   = Number of heterozygous individuals

_____   = Number of homozygous recessive individuals

Test how well your data fits the expected values from the equilibrium model:

_____ Chi-square test statistic

______ P value

_____ (y/n) in equilibrium?

4.3 In humans, the hitchhiker’s thumb trait is controlled by genes at a single locus, with the non-hitchhiker’s thumb allele dominant to the hitchhiker’s thumb allele. A population has 46 people that do not have the hitchhiker’s thumb and 21 that do. Of the humans without a hitchhiker’s thumb, 19 have a homozygous dominant genotype. Is this population in equilibrium?

Calculate p and q from the number of individuals of each genotype:

p = _____

q = _____

Calculate the expected frequency of each genotype if the population is in equilibrium:

_____   = Frequency of homozygous dominant individuals

_____   = Frequency of heterozygous individuals

_____   = Frequency of homozygous recessive individuals

Calculate the expected number of individuals of each genotype in a population of 67 humans if the gene is in equilibrium:

_____   = Number of homozygous dominant individuals

_____   = Number of heterozygous individuals

_____   = Number of homozygous recessive individuals

Test how well your data fits the expected values from the equilibrium model:

_____   Chi-square test statistic

______ P value

_____ (y/n) in equilibrium?

4.4 In a certain species of prickly pear, having straight or curved spines is a trait controlled by genes at a single locus, with the straight spine allele dominant to the curved spine allele. A population of prickly pears has 37 individuals with straight spines and 42 individuals with curved spines. Of the prickly pears with straight spines, 12 have a homozygous dominant genotype. Is this population in equilibrium?

Calculate p and q from the number of individuals of each genotype:

p = _____

q = _____

Calculate the expected frequency of each genotype if the population is in equilibrium:

_____   = Frequency of homozygous dominant individuals

_____   = Frequency of heterozygous individuals

_____   = Frequency of homozygous recessive individuals

Calculate the expected number of individuals of each genotype in a population of 79 prickly pear cacti if the gene is in equilibrium:

_____   = Number of homozygous dominant individuals

_____   = Number of heterozygous individuals

_____   = Number of homozygous recessive individuals

Test how well your data fits the expected values from the equilibrium model:

_____   Chi-square test statistic

______ P value

_____ (y/n) in equilibrium?

Answer 1

Answer 4.1:)

In the first population of birds:

Homozygous Red, YY = 9

Heterozygous Red, Yy = 22 – 9 = 13

Yellow, yy = 14

Total population = 36

Frequency of Homozygous Red, YY = 9/36

Frequency of Homozygous Red, YY = 0.25

Frequency of Heterozygous Red, 2Yy = 13/ 36

Frequency of Heterozygous Red, 2Yy = 0.36

Frequency of Homozygous Yellow, yy = 14/36

Frequency of Homozygous Yellow, yy = 0.39

Total alleles = 36 x 2

Total alleles = 72

Allele frequency of Y allele = ((9 x 2) + 13)/72

Allele frequency of Y allele = 31/72

Allele frequency of Y allele = 0.43

Allele frequency of y allele = ((14 x 2) + 13)/72

Allele frequency of y allele = 41/72

Allele frequency of y allele = 0.57

According to the Hardy-Weinberg equilibrium, the sum of allele frequencies must be 1 to ensure the population in equilibrium.

Sum of allele frequencies; Y + y = 0.57 + 0.43

Sum of allele frequencies; Y + y = 1

Therefore, the population is in equilibrium.

In the F2 generation, the expected genotype ratio should be 1:2:1, in which each genotype would get the following numbers:

YY = ¼ x 36

YY = 9

Yy = 2/4 x 36

Yy = 18

yy = 9

Chi-square test:

Progenies (b)	Observed (O)	Expected (E)	O-E	(O-E)2	(O-E)2 / E (Chi-square)
YY	9	9	0	0	0
Yy	13	18	-5	25	1.39
yy	14	9	+5	25	2.78
Total					4.17

Therefore, the chi-square value is 4.17. The degree of freedom is 2 for the 3 phenotypes.

The p-value of the chi-square value is 0.1, which is more than the significance value of 0.05. Therefore, the proposed hypothesis is rejected.

Based on the chi-square significance, the population is in equilibrium and there is no deviation.