meg
Forum Guru
Topics: 62
Posts: 806
|
Tay Sachs disease, an autosomal recessive disease, caused by a deficiency of hexosaminidase A, is lethal in childhood. In a population of Ashkenazi Jews, blood testing shows the frequency of heterozygotes to be 0.1. What is the probability that the first child of two individuals from this population with no family history of the disease will hae Tay Sachs?
A) 0.25
B)0.11
C)0.0625
D)0.0025
E) Cannot be determined from the information given
|
| guest
|
A) 0.25
|
| oddissy4u
Forum Guru
Topics: 107
Posts: 389
|
please explain the calculation.....
|
| Yulia
Forum Elite
Topics: 19
Posts: 240
|
0.25?!!!
Are you trying to say that 1 of 4 babies born to ashkenazi jew parents has tay sachs? No, it can't be so bad 
Here,
to have a baby with tay sachs both parents would have to be heterozygous, and frequency of heterozygotes is 0.1.
The chance that both of parents are heterozygous is 0.1 multiplied by 0.1 =0.01
and to calculate the probability of having an affected child, we need to multiply that number (0.01) by 0.25 which represents a recurrence risk for the offspring produced by 2 heterozygous parents).
0.01 multiplied by 0.25 = 0.0025 D is the correct answer.
|
| mjl1717
Forum Hero
Topics: 955
Posts: 5,451
|
Yulia I wrote it once before, but you Are the best! :!:
___________________
Smell the coffee! "Is That an Osler move??"
|
| Ouli Maty
Forum Elite
Topics: 33
Posts: 275
|
You are sure right about that. Yulia makes things sound so easy when in reality you struggle to even begun to understand them.
We are just grateful and happy to have her around.
___________________
deep breathing...
|
| robin082006
Forum Hero
Topics: 471
Posts: 5,125
|
The frequency of heterozygotes is 0.1, thus 2q=0.1 so q=0.05
The frequency of homozygotes is q*q = 0.05*0.05 = 0.0025
Because the first child and her/his parents are members of that population so Hardy Weinberg is applied for the calculation.
___________________
The Key to Succeed is Patience.
|
|
| |
| | | | | | | |
