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#1




Cystic Fibrosis Probability
Father is affected with CF phenotypicaly, mother is well phenotypicaly but her sister has cystic fibrosis. Came for prenatal checkup and want to know what chance of this child of this couple have cystic fibrosis?
a) 1/4 b) 2/3 c) 1/3 d) 3/4 
#2




probably 2/3
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#3




I feel like the answer would be 3/4 = 75%
I will be guessing these kinda of question on my exam.Hope I get them right. 
#4




oops i did mistake 1/3
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#5




C) 1/3
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riya rai (02282013) 
#6




a) 1/4...........

#7




1/3.. mother's sister had CF hence chances that the mother has the allele is 2/3. The father has the disease.. hence he is going to contribute anyways . hence the probability that the child has CF is equal to the probability that the mother carries the allele, which is 2/3 times the chance of getting a child which has the disease which is 1/2 hence the answer is 1/3

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riya rai (02282013) 
#8




answer c) 1/3
Quote:
ya you are corect mother has 2/3 chances of heterogenous mutant alele. 2/3*1/2 (2/3 times 1/2) = 1/3 this calculation is confusing esp for female doctrs pls break this calculation in simple way:sorry: 
#9




I have a confusion...
Quote:
The one here, though, is that when we say the mother's sister is affected, won't it mean the following options and not just option a) coz we don't know the genotype of the mother's parents and we have to satisfy the mother being heterozygous (carrier) and her sister being homozygous (affected) for recessive gene, which can be encountered in these two ways : a) both father and mother of the child's mother were heterozygous carriers and then the probability of mother being heterozygous is 2/3 b) one of the parent is homozygous for the recessive gene (affected) while the mother is heterozygous and then the probability of mother being heterozygous is 1 (100%) Why did we choose the a)? Am I missing something?? 
#10




You need to first calculate the odds that the mother has a defective allele. . Since her sister has the disease. . Both of her parents must be having the alleles. . Construct a diagram for recessive inheritance. . You can see that chances of disease is 1/4 , in the remainig three children. . Two are carriers and one is normal. . Hence the chance of the mother being a carrier is 2/3. . Does this help?

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riya rai (02282013) 
#11




"# The woman's parent is an obligate carrier coz they definitely got 1 allele from grandma, so her probability is 1
# The woman herself has a 50% chance of being a carrier from her parent, so her probability is 1/2 # Above two points are the same for the woman's husband as they have same grandparents # The probability of the heterozygous woman and her husband of having a child affected is 1/4 (autosomal recessive) so now multiply all these probabilities: 1/2 x 1/2 x 1/4 = 1/16" 
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Janakpriya (03012013) 
#12




It's 4) 1/256
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#13




Quote:

#14




In analogy to kaplan...
We need to find a heterozygote mate for the man (for one, heterozygotes are common in AR diseases, homozygote affected are rare and homozygote unaffected can't produce an affected child with this heterozygote man)
The question says that 1/40,000 of population is homozygous and we know from HardyWeinburg equation that 'homozygous population' is q^2 and recessive allele frequency (q) = 1/200 (underroot of 1/40,000) [I feel I need to be elaborate here coz at one point I was doing 1/6 * 1/6 = 1/12 ] Now we can calculate the probability of finding a heterozygous carrier from this population by again using the HardyWeinburg equation, which would be 2pq. Now we need p. Because p + q = 1, and we've calculated q from q^2 to be 1/200, therefore p = 1  1/200 = 199/200 ~= 1 So, coming back to what we need to find i.e., the probability of man finding a heterozygous carrier mate from the population which = 2pq and substituting the values calculated, 2pq = 2 * 1 * 1/200 = 1/100 Finally, we know that the man is heterozygous (given in question), so his probability of being heterozygous is 100% or 1/1 & We have calculated the probability of finding the man a heterozygous carrier from the population to be 1/100 & also probability of this hypothetical couple to produce an affected child = 1/4 (25%) (disease risk of child produced by these two heterozygous recessive couple from punnet square) NOW THE MOTHER OF ALL PROBABILITIES aka THE FINAL PROBABILITY = product of probabilities of each at the final stage = 1/1 * 1/100 * 1/4 = 1/400 This question made express emotions in the following sequence 
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sami trailer (03022013) 
#15




Quote:
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#16




A little confusion!
Quote:
Why can not we just say it in this way that, " if we assume the mother is heterozygous carrier for CF and father is homozygous recessive (affected) with CF, then there is 50% chance for a baby to get the disease. BUt if the mother is homozygous normal and father is homozygous diseased then there is 0% chance( all of the babies will be heterozygous carrier). why we are calculating the risk for mother?? And How we say that it is 2/3 chance the mother will be heterozygous as we don't know about her parents genotype! 
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