Homework Answers
Ans : Tay-sachs is a recessive disorder and is found in 1 in 3000 individuals every generation. Therefore this condition is in Hardy - Weinberg equilibrium.
Hardy-Weinberg equilibrium equation is,
p2 + 2pq + q2 = 1 and p + q = 1
where,
p2 = frequency of homozygous dominant genotype ("AA")
2pq = frequency of heterozygous genotype ("Aa")
q2 = frequency of homozygous recessive genotype ("aa")
p = frequency of dominant allele ("A")
q = frequency of recessive allele ("a")
a) Frequency of "aa" genotype = homozygous recessive genotype
Total individuals = 3000
Since Tay-Sachs disease is a recessive disorder (aa), the frequency of (aa) is represented as q2
Therefore,
q2 = 1 / 3000
q2 = 0.00033
b) Frequency of the q allele
We have calculated q2 = 0.00033
Therefore,
q = √0.00033
q = 0.0182
c) Frequency of p allele
We have equation ,
p + q = 1
Substituting q = 0.0182 in equation , we get
p + 0.0182 = 1
p = 1 - 0.0182
p = 0.9818
d) Carriers means the individuals have both the allele 'A' and 'a' in their genotype but is not expressing as it is a recessive disorder,
Therefore,
Frequency of individuals who are carriers for Tay-Sachs disorder = Frequency of heterozygous genotype (Aa)
Frequency of heterozygous genotype (Aa) = 2pq
We have p = 0.9818 and q = 0.0182
Substituting,
2pq = 2 x 0.9818 x 0.0182
= 0.0357
e) Frequency of homozygous dominant (AA) = p2
We have p = 0.9818
Therefore,
p2 = (0.9818)2
p2 = 0.9639
Final answer :
a) Frequency of "aa" genotype = q2 = 0.00033
b) Frequency of "q" allele = 0.0182
c) Frequency of "p" allele = 0.9818
d) Frequency of individuals who are "carriers" for Tay-Sachs = "2pq" = 0.0357
e) Frequency of "homozygous dominant (AA)" = " p2 " = 0.9639
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