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Author13 Posts
  #1

a man is a known heterozygous carrier for an autosomal recessive disease gene that has 60% penetrance in affected homozygotes. If he marries his second cousin, what is the probability that their offspring will be affected with the disorder (i.e., that they will manifest the disease phenotype)?

___________________
I hear and I forget. I see and I remember. I do and I understand.
--Confucius

  #2

generally,the offspring has 1/4th chance(25%) to be a homozygous recessive.and with disease penetrance of 60%,his probability to manifest the disease phenotype may be 1/4x0.6=0.15
do correct me please if i may be wrong.

  #3

but d ques asks d probability of his offspring having this ds if he marries his second cousin. So, taking dat into account it should be
1/4*1/32*0.6

___________________
I hear and I forget. I see and I remember. I do and I understand.
--Confucius

  #4

thanx mash for the correction

  #5

what is 1/4?
i suppose 1/32 is coeficient of relationship so why the answer isn't 1/32*1/2*0.6?

  #6

1/4 bcoz its an autosomal recessive disorder n 1/4 of d offsprings r going to be affected if both parents r heterozygous.....

___________________
I hear and I forget. I see and I remember. I do and I understand.
--Confucius

  #7

mash, are you sure this is the answer?
Is it a rule when talking about second cousin, the probability is 1/32?
Sorry I am bad in genetics, I get confused easily sad

  #8

Second cousins share a set of great grandparents. For any particular gene in the male, the chance that his female second cousin inherited the same gene from the same source is 1/16.
For any gene the man passes to his child, the chance is 1/16 that the woman has the same gene and ½ that she transmits that gene to the child so 1/16 X ½ = 1/32

so, chances of their offspring havin d ds r
1/32*1/4*0.6

i might be wrong so, plz correct me if u ve any other explanation.

___________________
I hear and I forget. I see and I remember. I do and I understand.
--Confucius

  #9

What I don't get here is the probability of the second cousin having the gene 1/16???how did you calculate it?

  #10

Yeah man! How come it's so complicated?

If I were to solve this problem, I'd think this way.

Ok, this guy is a known heterozygous. So he is Aa
His second cousin has a 1/2 of having the recessive gene b/c she looks normal. So she could be AA or Aa.

Now, for the offspring to show the disease, it's 1/4 chance he/she will be aa.
So, the chance for this kid to be aa is: 1/2 (for the mom to have the disease) X 1/4 (aa) X 60%

However I am not sure. Never took genetics except in med school!

___________________
La vita e bella!

  #11

actually it is a bit complicated stuff.........wat i know is
parents n children share 50% of their genes
siblings share 50%of their genes with each other
1st cousins share 1/8 of their genes
2nd cousins share 1/16 of their genes
coefficient of inbreeding is d probability dat the allele fr a trait inherited frm one parent is d same as inherited frm other parent if parents r related. (consanguinity)
so,in the above ques coefficient of inbreeding wud be 1/32

___________________
I hear and I forget. I see and I remember. I do and I understand.
--Confucius

  #12

Thnx for the explanation mash. So this is the officially accepted answer to ur q, correct? And if u're wrong, I'll come and hunt u down grin

___________________
La vita e bella!

  #13

absolutely bela , ans is 100% right..........
but im not sure of d explanation(that is why i ve quoted ''what i know " )
if smone has a better explanation he is most welcm......

___________________
I hear and I forget. I see and I remember. I do and I understand.
--Confucius







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