genetic 9 - basicq

[ArchivalUser]

albinism is an autosomal recesiive disease. The frequency of the gene for albinism in a population is know to be a 1/190. What is the aproximate risk for albinismin the child of an albino woman who marries an unrelated and unaffected man whit no family history of albinism?
a. 1/95
b. 1/190
c. 1/380
d. 1/570
e. 1/170

[ArchivalUser]

q=1/190

hence the hetrozygous carrier frequency in the population is 2pq=2*1/190=2/190=1/95

---probablity the man is a carrier=1/95
---probablity the man will pass the disease causing allele=1/2
---probablity the woman is an albino=1
---probablity the woman will pass the disease causing allele=1

probablity of independent events=1/95*1/2*1*1=1/190

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thank u basic for the questions.......!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

[ArchivalUser]

thanks yeabiruh for ur explanation......
i accept with ur ans 1/190

[ArchivalUser]

CC..

There probability that the mother will transmit the gene is 1
The probability that the father will have the gene is 1/190
the probability that the father will transmit the gene is 1/2
so the net probability of albinism in the child will be 1/380..


P.S: does the question say 'frequency of gene or is it frequency of allele'?.

[ArchivalUser]

answer is b
gene frequency = 1/190
so carrier frequency = 2 x 1/190
so chance of father being carrier is 2/190
mother is affected.. so chance of baby with albinism, for affected mother and carrier father is 1/2
so combined chance = 1/2 x 2/190 = 1/190


@sammy q asking abt baby with ablminism
@yeabiruh u welcome!