Question

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 127 people that do not have the hitchhiker’s thumb and 75 that do. Of the humans without a hitchhiker’s thumb, 42 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 202 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?

Answer 1

Answer:

Based on the given information:

• Non-Hitchhiker thumb allele is dominant to hitchhiker allele
• Let Non-Hitchhiker thumb allele be represented as H
• Let Hitchhiker thumb allele be represented as h
• Population size with Non-Hitchhiker thumb = 127 (dominant homozygous HH + heterozygous Hh)
• Humans without a hitchhiker’s thumb, number with homozygous dominant genotype (HH) = 42
• Heterozygous individuals population size (Hh) = 127-42 = 85
• Population size with Hitchhiker thumb (hh) = 75
• Total population size = 127 + 75 = 202
• Total number of alleles = 202*2 = 404
• Total number of h alleles = 75*2 + 85 = 235
• Total number of H alleles = 42*2 + 85 = 169
• Frequency of H alleles = 169/404 = 0.4183
• Frequency of h allele = 235/404 = 0.5817
• Genotype frequency for HH = 0.4183^2 = 0.1750
• Genotype frequency for hh = 0.5817^2 = 0.3384
• Genotype frequency for Hh = 2*0.4183*0.5817 = 0.4867
• P2 + q2 + 2pq = 0.1750 + 0.3384 +2*0.4183*0.5817 = 1.00

Chi square analysis:

	Observed (O)	Expected (,E)	(O-E)	(O-E)^2/E
HH	42	35.4	6.7	1.251
Hh	85	98.3	-13.3	1.803
hh	75	68.4	6.6	0.646
Total	202	202	Chi Square Sum	3.699

Degree of freedom = number of categories -1 = 3 - 1 = 2

Probability value corresponding to a chi square value of 3.699 with 2 degree of freedom is  P = 0.157316. The result is not significant at p < 0.05. Therefore, population is in Hardy Weinberg equilibrium.

Answers:

Calculate p and q from the number of individuals of each genotype: p = 0.4183 q = 0.5817

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

0.1750 = Frequency of homozygous dominant individuals

0.4867 = Frequency of heterozygous individuals

0.3384 = Frequency of homozygous recessive individuals

Calculate the expected number of individuals of each genotype in a population of 202 humans if the gene is in equilibrium:

36 = Number of homozygous dominant individuals

98 = Number of heterozygous individuals

68 = Number of homozygous recessive individuals

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

3.699 Chi-square test statistic 0.1573 P value Yes (y/n) in equilibrium?