theaero Forum Newbie
Topics: 4 Posts: 3
| | 05/13/05 - 01:55 AM  
 
   
 
|   #1 |
Three genes, D1, D2 and D3 are involved in ear formation. Homozygosity for a mutant allele (m) at any one of these three genes will cause deafness through the elimination of a critical component of the developing ear. In addition, homozygosity for all three genes causes prenatal lethality resulting in spontaneous abortion of the fetus. A married couple, both of whom are deaf, wish to have children. If the genotypes of the parents are as indicated below, what is the likelihood that their first child will have normal hearing? Mother: D1+/D1m, D2+/ D2m, D3m/D3m X Father: D1m/D1m, D2+/ D2m, D3+/D3m
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| son64 Forum Elite
Topics: 34 Posts: 245
| | 05/14/05 - 01:59 AM  
 
   
 
|   #2 |
0.1875
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| anjushree Forum Guru
Topics: 64 Posts: 386
| | 01/11/06 - 09:15 AM  
 
   
 
|   #3 |
ans-1/2 *3/4 *1/2 = 0.1875
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| msyamp Forum Fanatic
Topics: 60 Posts: 1,462
| | 01/11/06 - 01:28 PM  
 
   
 
|   #4 |
wow good one. anju you are great
___________________ If you think you can You can! If you think you cant you are right again!!
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| Geroo Forum Guru
Topics: 114 Posts: 799
| | 01/11/06 - 01:49 PM  
 
   
 
|   #5 |
agree with the answer
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| gogetit Forum Senior
Topics: 39 Posts: 104
| | 01/13/06 - 03:27 PM  
 
   
 
|   #6 |
can one of u explain to me pls... heres how i did it.... its given that Mother: D1+/D1m, D2+/ D2m, D3m/D3m Father: D1m/D1m, D2+/ D2m, D3+/D3m now take each allele and cross.... D1+/D1m X D1m/D1m, we get D1+D1m, D1mD1m, D1+D1m, D1mD1m... so chance of homozygosity is 1/2 D2+/ D2m X D2+/ D2m, we get D2+D2+, D2+D2m, D2+D2m, D2mD2m.... so chance of homozygosity is 1/4 D3m/D3m X D3+/D3m, we get D3mD3+, D3mD3m, D3mD3+, D3mD3m..... so chance of homozygosity is 1/2 So finally its 1/2 X 1/4 X 1/2 = 1/16.......
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| gogetit Forum Senior
Topics: 39 Posts: 104
| | 01/13/06 - 03:29 PM  
 
   
 
|   #7 |
and also what about the chance of prenatal lethality ??/ how to calculate that one?? i am lost.... pls help
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| msyamp Forum Fanatic
Topics: 60 Posts: 1,462
| | 01/13/06 - 03:35 PM  
 
   
 
|   #8 |
hey man. he is asking the normal listening child. so if you take hozygous for any allele he will be deaf. and more over you have caluculated hoozygous for all three alleles. so you got the probabilty of a dead fetus as said in the q. now you caluculate the heterozygous probabilty for each and multiply.you wil get probability of a health child. ok?
___________________ If you think you can You can! If you think you cant you are right again!!
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| gogetit Forum Senior
Topics: 39 Posts: 104
| | 01/15/06 - 08:11 AM  
 
   
 
|   #9 |
ok thanks mysamp... i got the heterozygus probability.... 1/2 X 3/4 X 1/2 = 3/16, but dont u need to subract 1/16.. for the chance of prenatal lethality...i am confused
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| pocas Forum Elite

Topics: 7 Posts: 169
| | 01/15/06 - 09:26 AM  
 
   
 
|   #10 |
the question ask yu about the health children no ask yu prenatal lethality, so focus in the heterozigotes children of each allele
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| msyamp Forum Fanatic
Topics: 60 Posts: 1,462
| | 01/15/06 - 09:26 AM  
 
   
 
|   #11 |
again substract? you are not counting the probability of all three homolugous diseased(prenataal lethal case) that is why you are getting 1/2 and 3/4 and 1/2
___________________ If you think you can You can! If you think you cant you are right again!!
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