probability que - aprna

[ArchivalUser]

a young couple present to your office for prenatal counselling,husband suffers from CF,and wife is carrier,what are the chances that couples unborn child will inherit CF
a-1/16
b-1/8
c-1/3
d-2/3

[ArchivalUser]

husband = disease = 100% Xc + Xc
wife = carrier = 50% Xx + Xc

Child 1: Xc + Xx = carrier
child 2: Xc + Xc = disease
child 3: Xc + Xx = carrier
child 4: Xc +Xc = disease

id say 50% but i dont see that option so i'd select D ???

[ArchivalUser]

ans -c 1/3

[ArchivalUser]

i thought same,but in answer it says 1/3,considering mothers inheritance,(chances of being carrier is 2/3),i am not getting this part.

[ArchivalUser]

where did u get this question from?

[ArchivalUser]

ahhhhhhhhhhhhhhh i see its going by hardy-weinberg equation:
p2+2pq+q2=1

p = normal homozygous
q = disease homozygous

so father = disease = q-q
mother = carrier = p-q
normal = p-p

thus child gettin affected = 1/3 .... q-q

[ArchivalUser]

you use the equation and then put the mother as 2pq and father q square

[ArchivalUser]

@zen786:its in qbank,thank you for explaining

[ArchivalUser]

oh okay!! and thank you aprna for a great question... always need a reminder or notes to look back to clearly states a tough question lol!!! thanks!

[ArchivalUser]

@zen786:i am almost mistaking all ques of probability,just need to get better at this concept.