








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




Three Genetics Questions
Case # 1:
What is the probability that the female carrier of an Xlinked disease will have a child with that disease assuming she mates with a normal male? Case # 2: Cystic fibrosis is autosomal recessive disease. Two parents that are heterozygous for cystic fibrosis have a normal unaffected child. What is the probability that the child is homozygous normal? Case # 3: If 49 %of a population is homozygous for curly hair gene that is dominant to straight hair gene, what percentage of the population has curly hair? Please give explanation. 
The above post was thanked by:  
#2




1  1/4
2 1/3 3 49 + 35 = 84 am i right? 
#3




1.1/4
2.1/3 3. P squared is 0.49.hence p=0.7(dominant allele) q=1p=0.3(recessive allele) q squared=0.09 hence 9% of popn has homozygous recessive and 1009= 91% has curved hair(homozygous dominant and heterozygous) Not sure if this is the answer. 
The above post was thanked by:  
aknz (12102011) 
#4




11/4th because only the male with affected x chromosome will have the disease,girls will be carrier
21/3rd since two heterozygous ppl mate in AR disease only one of the child will have the disease and other will be normal.Out of this 3 normal two are heterozygous 3since curly hair is dominant gene we use haardyweinberg equation here P SQUARE = 49 so P =7 now Q=1p=3 now total number of ppl in population with cury hairs equals P SQUARE + 2pq= 49+42=91% 
The above post was thanked by:  
#6




Have a question...
Quote:
1. Isn't that half of the daugthers will be carriers and half of the sons will be sick? So, isn't the answer 50%? (I'm not sure... I'm just asking) 2. Yes, I agree. One have to eliminate the homozygous to make the calculation. 3. At the begining I was wrong, then I did the calculation again and I think the same: 91%, becuase that will be the number of homozygous carriers for the gen of curly hair "p2" plus the heterozygous for the same gene "2pq". 
The above post was thanked by:  
aknz (12102011) 
#7




Quote:
Can you please explain me case 1?I did not understand it and fail to solve it. 
#8




Quote:
so think x2 is a mutant allele . . so what are the possible combinations XX,XY,X2X (carrier),X2Y(disease) so 1/4 
#9




Quote:
Thanks. 
#10




Quote:
Probability that she will pass the gene is 1/2 poababulty that she will hav a son is 1/2 so the probabilty that she will hav a child with disease is 1/2 * 1/2= 1/4!
__________________
Ppl come here to study n ponder I am here to Excell like Thunder! To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts. 
Tags 
Genetics, Step1Questions 
Thread Tools  Search this Thread 
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
If you don't die after this...you're done with genetics!  Claus_CU  USMLE Step 1 Forum  22  02202015 12:19 PM 
genetics  INCOGNITO  USMLE Step 1 Forum  4  08052011 10:10 PM 
Genetics questions  just_md  USMLE Step 1 Forum  0  06012010 07:37 PM 
Genetics of diabetes  Tsveta  USMLE Step 1 Forum  3  02152010 04:15 PM 
Genetics, not getting it well  Dr. N Elham  USMLE Step 1 Forum  3  12072009 07:45 PM 
