sobia naeem
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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
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| ssrpk
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if he mates with another heterozygote then the probability of hving homozygte affected wud be 25%
is it e???
i tried hardy-weinberg but cud'nt figure out!
plz explain!
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| ssrpk
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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
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| sobia naeem
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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!!!
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| ssrpk
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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
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| 99doc
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THE ANSWER IS B.0.045
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| ssrpk
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thnx a lot dude  
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| Geroo
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can some one explain the answer for this question please.this question is in kaplan book and in q bank with two different answers
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| Doc2378
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Huh?...2 answers from the same publisher!...extremely strange..
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| Geroo
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yep,with 2 different explanations.did anyone notice that??
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| sturge_weber
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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
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| sturge_weber
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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
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| roopashri
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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 aa---Aa, 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 ...
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| sturge_weber
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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
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| Geroo
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thanks sturge and roopashri
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| roopashri
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| drk1980
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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)??
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| roopashri
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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
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| drk1980
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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.
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| roopashri
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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
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| drk1980
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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!)
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| drk1980
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thanks to you for ur input
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| cyra
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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!!!
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| cyra
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Guess you guys figured it out...the father's carrier state has to be a 1 i.e a 100% because we know he is the carrier!!
Good job dr k!
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| Geroo
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I think it's impossible for anyone to answer such a question if it shows up in the exam,that's really a hard question,at least for me.even kaplan guys answered it wrong in the book then they modified the answer in qbank.
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