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Q.   In a random mating population frequency of disease causing recessive allele 80%. What would be the frequency of carrier individual ?

                  q = 80%
                  q = 0.8
                  p+q = 1
                  p = 0.2

Dear student,                                                                                                                                
​Please find below the solution to the asked query    

According to Hardy-Weinbergs Principle: 

p 2  + 2pq + q 2   = 1 

where:

   p2  =frequency of occurrence of healthy individuals 

  q2 =frequency of occurrence of diseased individuals

2pq = frequency of occurrence of carrier individuals 

Given: 
q = 0.8

Since, 
p + q = 1 
p = 1 − q 
  = 1 − 0.8
  = 0.2

Frequency of carrier individuals= 2pq
                                                    = 2 x 0.2 x 0.8
                                                   = 0.32

Hence, the frequency of occurrence of carrier population is 0.32 or 32 %.

Hope this information will clear your doubts about the topic.    Regards                                                                                              

Hope this information will clear your doubts about the topic.    Regards                                                          ​