BI0349 FA18 Ch. 25 HW Problems Name: 1. In a population of 2000 Gaboon vipers (B
ID: 141171 • Letter: B
Question
BI0349 FA18 Ch. 25 HW Problems Name: 1. In a population of 2000 Gaboon vipers (Bitis gabonica), a genetic difference with respect to venom exists ata single locus that controls venom production. The alleles are incompletely dominant. The population shows 150 individuals homozygous for the t allele (genotype tt, non-venomous), 750 heterozygous (genotype Tt, mildly venomous), and 1100 homozygous for the T allele (genotype TT, lethally venomous). (3 pt) a. What is the frequency of the t allele in the population? (Show Work) (2 places after decimal) b. Are the genotypes in Hardy-Weinberg equilibrium? Why or why not? (Show work) HINT: Use Chi-Square Table and Calculate Degrees of freedom as (# Genotypes-# Alleles Possible)-DF. Why might they not be in H-W equilibrium? What kind of selection pressure could be at play?Explanation / Answer
1) a) The population of gaboon vipers is 2000. Out of 2000, 150 were homozygous for t allele. 750 were heterozygous Tt and the rest 1100 were homozygous for T allele TT.
The frequency of t allele,
= (150 + (750/2)) / 2000
= (150 + 375) / 2000
= 525 / 2000
= 0.2625
B) The frequency of T allele is,
= (1100 + (750/2))/ 2000
= (1100 + 375) / 2000
= 1475/2000
= 0.7375
If the population is in Hardy Weinberg's equilibrium,
p^2 + 2pq + q^2 =1
= (0.7375)(0.7375) + 2(0.7375)(0.2625) + (0.2625)(0.2625)
= 0.544 + 0.387 + 0.069
= 1
The number of TT individuals should be = (0.544)(2000) = 1088
The frequency of heterozygous individuals should be = (0.387)(2000) = 774
The frequency of tt individuals = (0.069)(2000) = 138
Chi square test is done as follows,
Degree of freedom = number of observations - 1 = (3-1) = 2
For degree of freedom 2 and significance level of 0.05, the table value is 5.99
The obtained chi square value 1.919 is less than the critical value of 5.99.
So, the genotypes are not in Hardy Weinberg's equilibrium.
C) They are not in H-W equilibrium because of directional selection as there are more number of TT individuals.
Genotype O E O-E (O-E)^2 (O-E)^2/E TT 1100 1088 12 144 0.132 Tt 750 774 -24 576 0.744 tt 150 138 12 144 1.043 Total 1.919Related Questions
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