sturge_weber Forum Guru
Topics: 77 Posts: 1,042
| | 12/22/05 - 03:51 PM  
 
   
 
|   #26 |
well geroo i thinkthe exam if they give a choice like 0.045, then mark it, if they dont give the first choice ie 0.045 the mark the second one, ie after calculating the womans probability with disease, and if both given then mark the q bank one one, this is just an opinion, u will do this q right if u get on the exam.... believe me 
|
| Geroo Forum Guru
Topics: 114 Posts: 799
| | 12/22/05 - 04:31 PM  
 
   
 
|   #27 |
yes sturg if i get the exact same question I will answer it right we know the answe already,but if we havn't seen it b4 and we get it in the exam it's kinda hard to figure it out and do the calculations
|
| sturge_weber Forum Guru
Topics: 77 Posts: 1,042
| | 12/22/05 - 04:50 PM  
 
   
 
|   #28 |
even if differnet, values, but same type. u will do it right, believe in urself now
|
| Geroo Forum Guru
Topics: 114 Posts: 799
| | 12/22/05 - 05:03 PM  
 
   
 
|   #29 |
ha ha thanks sturge
|
| satyaking Forum Junior
Topics: 5 Posts: 58
| | 12/24/05 - 12:26 AM  
 
   
 
|   #30 |
well guys, i find the answer 0.045 is not correct,because he is asking probability between man who is a heterozygote that we knew and woman we donot know what is her genotype and probability is. so when a cross between heterozygote man and heterozygote woman in population is 1 . 0.09 . 1/4 that is equal to 0.09/4 probability of affected child (we know man is heterozygous but womans probability we dont know it is pq that is 0.09) and whena cross occurs between male heterozygote and homozygous female the probability of child being affected is 1 . 0.01 . 1/2 = 0.01/2 hence by adding the probability from both possibilities that is 0.09/4 + 0.01/2 we get answer 0.025,hence it seems to me 0.025 is the answer,if iam wrong please correct me
|
|
| |
| | | | | |