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 Tay-Sachs disease  




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Author8 Posts
  #1

A woman is planning to marry her first cousin, but the people discovers that their shared grandfather’s sister died in infancy of Tay-Sachs disease.
What is the probability that the cousin’s first child have Tay-Sachs disease, assuming that all people who marry into the family are homozygous normal?





  #2

why you, ladies and gentlemen, did not answer? You don't like a test... raised eyebrow

ok these are possible answers:

a. 1/2

b. 3/4

c. 1/144

d. 1/96


  #3

OlgaG wrote:
why you, ladies and gentlemen, did not answer? You don't like a test... raised eyebrow



cause it is very very time consuming
i awlays get these Qs wrong on first thinking
if i got on of these in my exam i will choose the second rarest probability without thinking grin and save my time and my brain

but because we are not in exam i think the answer is d. 1/96

how i get it :
u have to calculate 3 probabilities for an autosomal recessive gene

1st ; the probability that the grandfather is heterozygous for the disease ?
it 2/3 why because his phenotype is normal
2nd : the probability that both mother and father has the mutated gene
is 1/4 x 1/4
3rd ; probability that their child will be homozygous for tay sachs = 1/4

so the sum = 2/3 x 1/4 x 1/4 x 1/4 = 1/96
right ???raised eyebrow


  #4

The chance that grandfather is heterozygous A/a is 2/3

The chance that each of parent is heterozygous A/a is 2/3*1/2=1/3

The chance that each of cousin is heterozygous A/a is 1/3*1/2=1/6

Chance that offspring is a/a with disease is 1/6*1/6*1/4=1/144

I agree that this is time consuming but our goal to make a lot of these tests and on the real exam do it fast and earn additional points.

First i also answered 1/96 but the real answer is 1/144.


  #5

ok but there is something i don't understand

they share the same grandfather so u only need to calculate his risk only once

in statistical words : the grandfather probability is mutually exclusive for both events (if it happen to mother it will also happen to the father of the child )

i mean that the grandfather probability to carry the mutated gene is one event for both mother and father so why do we need to calculate this probability twice

do i miss something ?raised eyebrow


  #6

I think we do it because if grandfather A/a , his daughter can be A/a or A/A, his son can be A/A or A/a, right?

Their children also can be A/a or A/A, that is why we need to calculate probability for each parent separately.

2/3*1/2*1/2=1/6 one parent

2/3*1/2*1/2=1/6 another parent

1/4*1/6*1/6=1/144 child


  #7

Hi There ,
That's very goog Qs.
My Brain is starting to work,
Waiting for other same Qs from you.........


  #8

even i think its 1/96






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