Respuesta :
Answer:
Option B
Explanation:
Given
Number of incidences or frequency of an autosomal recessive disease [tex]= \frac{1}{10000}[/tex]
As per Hardy Weinberg's equation, frequency of a recessive genotype is [tex]q^{2}[/tex]
[tex]q = \sqrt{\frac{1}{10000} }\\ q = \frac{1}{100}[/tex]
As per first equation of Hardy Weinberg's -
[tex]p+q=1[/tex]
so ,
[tex]p = 1-q\\p = 1-\frac{1}{100}\\p = \frac{99}{100}[/tex]
[tex]p^2 = (\frac{99}{100})^2\\p^2 =\frac{9801}{1000}[/tex]
As per second equation of Hardy Weinberg's -
[tex]p^{2} + q^{2} + 2pq=1[/tex]
Substituting the given values in above equation, we get -
[tex]\frac{9801}{10000} + \frac{1}{100} + 2pq=1\\2pq = 1-(\frac{9801}{10000} + \frac{1}{100})\\2pq = 0.99%\\
OR
[tex]2pq=1[/tex]%
[tex]2pq = \frac{1}{100}[/tex]
Hence, option B is correct.
