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  #1  
Old 02-27-2013
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Genetics 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
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  #2  
Old 02-27-2013
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probably 2/3
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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.
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oops i did mistake 1/3
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C) 1/3
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a) 1/4...........
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Old 02-27-2013
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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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Default answer c) 1/3

Quote:
Originally Posted by 4u.venkatesh View Post
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
answer is c) 1/3
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:
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Old 02-28-2013
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Default I have a confusion...

Quote:
Originally Posted by 4u.venkatesh View Post
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
Not one but lotsa

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??
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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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"# 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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It's 4) 1/256
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Quote:
Originally Posted by singular View Post
So, this is what I was missing!! Thanks! Thanks! Need to keep that in mind.




There's a question from KLN which is haunting me.. Here it goes:

A man is known heterozygous carrier of a mutation causing hyperprolinemia, and autosomal recessive condition. Phenotypic expression is variable and ranges from high level of excretion of proline to neurological signs and symptoms. Suppose that 0.0025% (1/40,000) of population is homozygous for the mutation causing this disease. If the man mates with somebody from population, what is the probability that he and his mate will produce a child who is homozygous for this condition..

a) 1% (1/100)
b) 0.5%(1/200)
c) 0.25%(1/400)
d) 0.1%(1/1000)
e) 0.05% (1/2000)


It has been asked before but didn't answer what I was looking for.

The issue is why don't we include the probability of the man mating with a female as well while calculating the final probability (others including the mate to be a carrier 1/100, the mate passing the mutant allele 1/2, and the man also passing the mutant allele 1/2)? Obviously a male won't mate with another but don't we have to include another probability of 1/2 (50%) of finding a female from the population? Just because it's all being done in probability? Kaplan doesn't do that..

Am I unnecessarily overthinking? Or have I gone nuts I feel I'll mess up all similar questions on the exam. What would you suggest to improve ? It's my 1st read.
can you please post the answer/explanation please :sorry:
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Default 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 Hardy-Weinburg equation that 'homozygous population' is q^2 and recessive allele frequency (q) = 1/200 (under-root 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 Hardy-Weinburg 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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Old 03-01-2013
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Quote:
Originally Posted by singular View Post
"# 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"
Thanks singular.... looks simple now...)
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  #16  
Old 03-07-2013
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Warning! A little confusion!

Quote:
Originally Posted by riya rai View Post
father is afected ( homogenous mutant allele 1/4) aa,
mother is carier (hetero mutant alele 2/3) Aa,
so child 2child wil be hetero mutant alele Aa,
and 2 other child wil be homo mutant allele aa,
so acording to pedigree we get this,am i right?
now chances of afected child wil be 1/2 (homo mutant alele).
mother has 2/3 chances of inherit the mutant alele to kid and kid has 1/2 chances of geting disease from both parents.
now 2/3 times 1/2 =1/3.
am i right?

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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