Prep for USMLEPrep for USMLE
         Forum      |     Resources New Posts   |   Register   |   Login





 q  



Post Reply  
  • 0/5
  • 1
  • 2
  • 3
  • 4
  • 5


Author14 Posts
  #1

A man who has Neurofibromatosis type 1 (autosomal dominant) marries a phenotypically normal woman. If they have five children, what is the probability that none of the children will be affected with this disorder? What is the probability that all of the children will be affected?




  #2

both 1/32.


  #3

both... 50%..???


  #4

both 1/32.but how?????can u plz xplain it drgho
thanq


  #5

i guess i got it..chances are 50% that is 1/2 and as he has 5 children so probability is 1/32 but still plz do explain it...


  #6

ya i got it hope 4 dbest.it z d chance of gettin d disease is 1/2.five children having d disease is (1/2)*5.b cos dey r independent events we multiply.same eg. given by daughtery in lecture chance of gettin tails in succesion is 0.5*0.5.am i right drgho???


  #7

since dis is an autsomal dominant,we xpress it r not xpress it.d chance of xpressing it is 1/2.


  #8

hope4dabest wrote:
i guess i got it..chances are 50% that is 1/2 and as he has 5 children so probability is 1/32 but still plz do explain it...


Sorry for late reply. The probability of having one child affected for that couple will be 1/2. So, the probability of having all 5 children affected will be 1/2 * 1/2 * 1/2 * 1/2 * 1/2 which is 1/32.
That is the application of binomial equation.
Let's say this couple have 3 children and what is the probability of having 2 children unaffected and one children affected?
smiling face


  #9

unaffected---1/2*1/2=1/4

affected--1/2

did i get it?????



  #10

hope4dabest wrote:
unaffected---1/2*1/2=1/4

affected--1/2

did i get it?????

shaking head...only one probability... not two probabilities separated.


  #11

drgho wrote:

shaking head...only one probability... not two probabilities separated.


but there r 3 children in this case and chance of expressing this gene is 1/2 and that also means not expressing is the left out that is 1/2....and as no of children are independent events so out of 3 children, probability 0f 2 children for not affected wont be multiplying 2 times and and for affected.........

i know i am in deep shit...plzzz explain again???




  #12

or do u mean probability for 3 children of being affected or unaffected wil be the same that is 1/2*1/2*1/2


  #13

hope4dabest wrote:
or do u mean probability for 3 children of being affected or unaffected wil be the same that is 1/2*1/2*1/2


yep..it is 1/8 but...there are also three combinations to have one affected and two unaffected. Out of three children named a, b and c. It is possible that Abc or aBc or abC (capital letter for affected one). So the probability is 3/8.



  #14

ok..thank u sooo much....i will try to look some more questions of these types....





Bookmark and Share



This thread is closed, so you cannot post a reply.



Login or Register to post messages








show Similar forum topics

show Related resources









Advertise | Support | Premium | Contact