Prep for USMLE 
Forum  Resources  New Posts  Register  Login  » 

Author  24 Posts  
sobia naeem
Forum Elite Topics: 30 Posts: 303 
a man is a known heterozygous carrier of a mutation that causes hemochromatosis(autosomal recessive)..suppose that 1% of the general population consists of homozygotes for this mutation.if the man mates with somebody from the general population what is the probability that he and his mate will produce a child who is an affected homozygote?? A)0.025 B)0.045 C)0.09 D)0.10 E)0.25  
Prep4USMLE.com
Advertisement 


ssrpk
Forum Junior Topics: 154 Posts: 2850 
if he mates with another heterozygote then the probability of hving homozygte affected wud be 25% is it e??? i tried hardyweinberg but cud'nt figure out! plz explain!  
ssrpk
Forum Junior Topics: 154 Posts: 2850 
p2 +2pq + q2 if q2 is 0.01 then q=0.1 and 2pq will be 0.2 (approx) if he mates with a heterozygote then tyhe probaility of hving an affected child (considering the whole general population) shud be 2pq X 2pq [jointr probabilituies of independent events] or 0.2 X 0.2= 0.04 i'll go with B  
sobia naeem
Forum Elite Topics: 30 Posts: 303 
well ssrpk this question is from kaplan lecture notes pg 307..i didnt have the answer printed so i thought i,d ask someone..i cudnt work it out myself!!!  
ssrpk
Forum Junior Topics: 154 Posts: 2850 
haha damn we are in the same boat..... i am missing tht page too n answers for q3,4,5,6&7.... somebody plz post the right answer for this q  
99doc
Forum Elite Topics: 52 Posts: 370 
THE ANSWER IS B.0.045  
ssrpk
Forum Junior Topics: 154 Posts: 2850 
thnx a lot dude  
Doc2378
Forum Guru Topics: 46 Posts: 678 
Huh?...2 answers from the same publisher!...extremely strange..  
sturge_weber
Forum Guru Topics: 77 Posts: 1034 
i didnt, but can u explain what the other ans is, my ans is 0.045, q2=1/100, q=1/10, 2pq=2x1/10x9/10=.18 0.18x1/4= 0.045 can u explain the other explanation q bank has given  
sturge_weber
Forum Guru Topics: 77 Posts: 1034 
well ya i checked the qbank , u re right, the second choice he has calculated till 0.045 and then he says theire will be an additional chance that the person he mates with is an autosomal recessive and has the disease... which is still not clear, please if someone know this explain how he calculated 0.01/2.... good job geroo, i still remember doing the q and looking for 0.045.. ha ha there was no choice and i just guessed a wrong ans, good job geroo  
roopashri
Forum Senior Topics: 16 Posts: 189 
thanks geroo for pointing it out... well i think they also dint know it when they published it in qbook... so they rectified in qbank ... ahaha sturge, after getting 0.045, u need to calculate the additional chance of the lady being homozygote, which is given as 0.01 hemochromatosis is autosomal recessive, so aa will be the homozygote for the disease. Aa x aaAa, Aa, aa, aa,Therefore 1/2 is the chance the man and a homozygous dieased woman will produce a child as a carrier. so its0.01 X 1/2... hope i dint confuse u ...  
sturge_weber
Forum Guru Topics: 77 Posts: 1034 
o ya i got u, good , thanks for clearing my doubt....so 1/2 is thee chance they will produce the affected homozygote yep thanks roo, as mjl1717 calls u  
roopashri
Forum Senior Topics: 16 Posts: 189 
 
drk1980
Forum Guru Topics: 147 Posts: 1036 
sorry ppl, i still had a doubt in this one(hope i dont get a single Genetics calculation on my exam)!! sturge, in your calculation, you used 2pq(carrier frequency) only once.....for one of the parents....wht abt the other parent? dont u take the carrier frequency of both parents into account in a mating? so 2pq * 2pq * risk of recurrence(ie. 1/4)??  
roopashri
Forum Senior Topics: 16 Posts: 189 
drk1980, read the qn pls... its given that man is a carrier, we should need to know the probability of the woman... hope u got it  
drk1980
Forum Guru Topics: 147 Posts: 1036 
roopashri, i think we already calculated the probability of the woman being a carrier with 2pq. the Q asks for the probability of the child being affected after mating between the two.  
roopashri
Forum Senior Topics: 16 Posts: 189 
drk, the qn stem gives us that man is a carrier right? we calculated the chance of woman being carrier as 2pq=0.18 Now these 2 carriers mate and the chance of them having an affected child is 0.25. So, the prob is 0.18 X 0.25=0.045. then there can an additional chance that the woman can be homozygote, which is given in qn stem as 0.01. So when a carrier man and homozygote woman mate, 1/2 is the chance that they will produce a child as a carrier. so its0.01 X 1/2=0.005 now final answer is 0.045+0.005=0.05 let me know if i am wrong anywhere, i will correct myself.. or if iam right then am sorry for being bad at teaching  
drk1980
Forum Guru Topics: 147 Posts: 1036 
i think i may hv figured this out..... contrary to wht i thot, tht we werent considering father's carrier state in the calculating offsprings homozygous affected probability[recall: 2pq(probability of mother's carrier) * 1/4= answer] The father's carrier state is also multiplied in this equation; its just that it is 100% or 1, so it doesnt add to the numbers(shd be written: 1*2pq*1/4=answer). (hope my logic is right!)  
drk1980
Forum Guru Topics: 147 Posts: 1036 
thanks to you for ur input  
cyra
Forum Guru Topics: 29 Posts: 843 
You know what drk....you do have a point!!! We need to figure out the probability of the man and the woman mating and producing an affected homozygote.... 2pq*2pq*0.5 ..... Add to this the probability that the man mates with a homozygote woman and produce a homozygote child which is the following... 2pq*q2*0.5(probability of producing a homozygote child) Pls do the math for me guys...shall check in on this later... I hope to God i dont get any questions like these...i am awful with numbers as it is!!!  
This thread is closed, so you cannot post a reply.
Advertise  Support  Premium  Contact 