1. In a population of 2000 Gaboon vipers (Bitis gabonica), a genetic difference
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Question
1. In a population of 2000 Gaboon vipers (Bitis gabonica), a genetic difference with respect to venom exists at a 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 Tallele (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.Explanation / Answer
Phenotype
Lethally Venomous
Mildly Venomous
Non venomous
Total
Genotype
TT
Tt
tt
Number of individuals
1100
750
150
2000
Observed genotype frequencies
1100/2000=0.55
750/2000=0.38
150/2000=0.08
=1.0
Allele frequencies
p=0.74
q= 0.26
=1.0
Expected genotype frequencies
0.54
0.39
0.07
=1.0
For Allele Frequencies:
There are two T alleles in TT = 1100 * 2= 2200
There are one T alleles in Tt= 750
Total number of alleles= 2* 1000+2* 750+2*150=4000
Frequency of allele T = p= Total number of alleles that are dominant/Total number of alleles= 2200+750/2000+2000= = 0.7375
Frequency of allele t = q= 1-p= 1-0.7375=0.2625
For Expected Frequencies:
The Hardy Weinberg equation is given as:
p2+2pq+q2=1
p2 = frequency of the homozygous genotype TT, q2 = the frequency of the homozygous genotype tt, and 2pq = frequency of the heterozygous genotype Tt.
Expected frequency of TT= p2= 0.7375*0.7375= 0.5439
Expected frequency of Tt= 2pq= 2X 0.7375 X 0.2625= 0.3872
Expected frequency of tt= q2= 0.2625 X 0.2625= 0.0689
Question a:
Frequency of allele T = p= 2200+750/2000+2000= = 0.7375= 0.74
Frequency of allele t = q= 1-p= 1-0.74=0.26
Frequency of t allele in population =q= 0.26
Question b:
Chi-Square test:
Null Hypothesis: the frequencies are in Hardy-Weinberg equilibrium
Observed
Expected
(O-E)
(O-E)2
(O-E)2/E
TT
0.55
0.54
0.01
0.0001
0.000185
Tt
0.38
0.39
-0.01
0.0001
0.000256
tt
0.08
0.07
0.01
1E-04
0.001429
Total
0.9997
1
0.01
0.0003
0.00187
Degree of freedom= Number of genotypes-Number of alleles= 3-2=1
Probability of chi square at d.f 2= 0.89= 89%
A Chi Square value between 1%-5% should allow you to reject the null hypothesis. However, 89% is a large P Value. Hence, the null hypothesis is accepted. The frequencies are in Hardy-Weinberg equilibrium.
Phenotype
Lethally Venomous
Mildly Venomous
Non venomous
Total
Genotype
TT
Tt
tt
Number of individuals
1100
750
150
2000
Observed genotype frequencies
1100/2000=0.55
750/2000=0.38
150/2000=0.08
=1.0
Allele frequencies
p=0.74
q= 0.26
=1.0
Expected genotype frequencies
0.54
0.39
0.07
=1.0
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