








USMLE Step 1 Forum USMLE Step 1 Discussion Forum: Let's talk about anything related to USMLE Step 1 exam 

Thread Tools  Search this Thread  Display Modes 
#1




Autosomal dominant disorder and normal offspring
A man and woman are both affected by an autosomal dominant disorder that has 80% penetrance. They are both heterozygotes for the diseasecausing mutation. What is the probability that they will produce phenotypically normal offspring?
A) 20% B) 25% C) 40% D) 60% E) 80% 


#4




If you work out a punnett square then the chance of a normal child is 25%.
But Im confused about the 80% penetrance. Does that mean you subtract 20% (10080) of 25% ? .. so 25% x 0.2 = 5%. Therefore the probability to have a normal child will be 20%. 
The above post was thanked by:  
bebix (07122011) 
#6




Quote:
but offspings' chance of having disease = 75 %!!!... phenotypically chance of having disease = 75 % x 80% =0.6 
#7




chances of offspring to carry abnormal genes is 75%
if the abnormal gene does not express itself, then there is no disease manifestations. so when you say disease, it depends on whether you mean genotype or phenotype 
#8




80% x 1/4 ( which represent the normal phenotype ) = 20% ...
Im not good at this,, i just try...  A  a AA A  Aa aA a aa This disease is DOMINANT which means just 1 alele is require to express the disease so 3/4 are Sicks 1/4 is normal... Please if im wrong explain it... =) So According to my plots Answer is A 
#9




Quote:
If both parent are heterozygotes, there is a 75% chance that their offspring will receive one or two copies of the diseasecausing gene. With 80% penetrance, the probability that the offspring will be affected (phenotypically) is 0.75 * 0.8 = 60%. The probability that the offspring will be phenotypically normal is 1  0.6 = 0.4 or 40% 
#10




Thanks Bebix....=)

#11




Quote:

The above post was thanked by:  
drgsarunprasath (07122011) 
#12




Quote:

#13




two ways to answer this question:
Aa x Aa = AA Aa Aa aa 80% penetrance a) 0.75 * 0.80 = 60% affected (phenotypically) 10060 = 40% "normal" b) 0.25 (aa) * [0.75 * 0.2] where, [0.75 * 0.2] = phenotypically normal = 40% 
The above post was thanked by:  
drgsarunprasath (07122011), drortho (11252011), Hope2Pass (07122011), monsalg2502 (08242011) 
Tags 
Genetics, Step1Questions 
Thread Tools  Search this Thread 
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
List of Autosomal Dominant Diseases  rasheed  USMLE Step 1 Bits & Pieces  11  05302013 04:34 PM 
Autosomal Dominant Inheritance Question  Claus_CU  USMLE Step 1 Forum  8  06072011 08:45 PM 
Conduct Disorder Vz Antisocial Personality Disorder  Ayshee  USMLE Step 2 CK Mnemonics  1  10302010 11:33 AM 
The difference between Conduct Disorder and Oppositional Defiant Disorder  TelsaFarad  USMLE Step 2 CK Forum  2  06152010 03:08 PM 
Delusional Disorder or Brief Pshychotic Disorder  obgynaim  USMLE Step 1 Forum  1  12202009 09:06 AM 
