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#1
01-24-2012
 USMLE Forums Addict Steps History: Step 1 Only Posts: 177 Threads: 33 Thanked 86 Times in 16 Posts
Calculating the probability of genetic diseases

Can somebody explain the way probability should be calculated in different sets of diseases with different inheritance patterns.
Please with thorough examples to understand explanations.

Would be very thankful.

Thanks
 The above post was thanked by: LatinGeorge (01-24-2012)

#2
01-25-2012
 USMLE Forums Addict Steps History: Step 1 Only Posts: 136 Threads: 18 Thanked 84 Times in 37 Posts Reputation: 94

hi poonam,, let me try.....

AUTOSOMAL RECESSIVE
if both the parents are carrier of autosomal recessive disease, the probability of the child to become a carrier would be 2/4 and for the disesased would be 1/4.

Mother is carrier(Pp), Father is also carrier(Pp),,so the possible conbinations are: PP,Pp,pP,pp.
Here capital P is disease causing gene. As we are talking about autosomal recessive disease, person having PP will be affected, while having one P will be carrier.

Therefor, 1 out of 4 will be diseased,so probability would be 1/4.
2 out of 4 will be carrier, so probability is 2/4.

AUTOSOMAL DOMINANT
Here having a single disease causing gene will produce disease.
Let say, mother is affected(two possibilities: she might be ZZ or Zz) and father is not(zz)

possible combinations for ZZ mother(affected) and zz father(normal) would be: Zz,Zz,Zz,Zz.
All four have capital Z. so all will have disease.4/4=1

possible conbination for Zz mother(affected) and zz father(normal): Zz,Zz,zz,zz.
2 out for have Z. so 2/4= 1/2.

possible combination for Zz mother(afffected) and Zz father(affected): ZZ,Zz,zZ,zz.
3 out of 4 will have Z. so 3/4

X linked recessive

females are mostly carriers and males are affected.

let say male is XY(normal) and female is XX.(carrier, X-disease causing gene)

possible combinations would be XX,XX,XY,XY.

Here, only one male(XY) will be affected as he has only one x chromosome which is affected. so probability of the diseased child would be 1/4.

One female would be carrier(XX). probability would be 1/4.

Let say,male is afffected (XY) and female is normal(XX).

Possible conbinations would be XX,XX,XY,XY

no one would be affected,
carrier probability 2/4= 1/2.

Let say man is affected(XY) and female is carrier(XX)

possibilities would be XX,XX,XY,XY.

here the possibilities for the disease is XY and XX ,so 2/4= 1/2..
Carrier posibilities is XX only.. so its 1/4.

Hope you understood........its hard to describe this way..i tried my best....good luck
 The above post was thanked by: bunny (01-25-2012), Dr. Mexito (03-24-2012), Hitman (01-25-2012), Poonam aslam (01-25-2012)
#3
01-25-2012
 USMLE Forums Addict Steps History: Step 1 Only Posts: 177 Threads: 33 Thanked 86 Times in 16 Posts

Thanks...I have to start genetics from tomorrow....ya it's a great great help...but pz pz don't mind one or two more questions...will be a lot of help.thanks thanks thanks again.

#4
01-26-2012
 USMLE Forums Scout Steps History: Not yet Posts: 28 Threads: 1 Thanked 10 Times in 8 Posts Reputation: 20

Brilliant answer. I just wanted to add to add to the discussion:

From what I understand, where there are NO risk factors (ie- neither parent has it), we use the Hardy Weinberg equation
p2+2pq+q2

The small 2s are supposed to be squares. q is the chance of having the disease, and p is of not.
So, q2= Incidence in the general population.
 The above post was thanked by: riya rai (02-27-2013)

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