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Using the Hardy-Weinberg Equilibrium, it is possible to establish the ratio of dominant to recessive alleles.
P represents the dominant allele.
q represents the recessive allele.
|Parents:||Aa x Aa|
|Gametes:||A or a||A or a|
Applying the P / q
P2 + 2Pq + q2
100% of the population will be AA, Aa or aa
Therefore: P2 + 2Pq + q2 = 1
The total number of dominant alleles + The total number of recessive alleles = 100%
P + q = 1
A genetic disease is caused by a recessive allele. The frequency of the disease in the population is 1 in 2,000. Calculate the frequency of the carrier genotype.
Let P = normal, q = disease allele.
NN = normal (P2),
Nn = carrier (2Pq),
nn = sufferer (q2).
q2 = 1 in 2,000 = 0.0005
q = √0.0005 = 0.0224
Since P + q = 1
P = 1 - q = 1 - 0.0224
P = 0.9776
2Pq = 2 (0.9776 x 0.0224) = 0.044
In 23 people (roughly 5% of the population are carriers).
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