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#1
05-28-2011
 USMLE Forums Addict Steps History: 1 + CS Posts: 113 Threads: 8 Thanked 89 Times in 61 Posts Reputation: 99
Genetic Counseling of CF Parents

Cystic fibrosis shows autosomal recessive inheritance. Parents of a newly diagnosed affected child are referred for genetic counseling. It would be correct to tell them that:

a) the probability that their next child will be affected is 1 in 2.
b) the probability that the older unaffected sister of the affected child is a carrier is 1 in 2.
c) the fact that their last child was affected means that their next three children will not be affected.
d) the probability that each parent is a carrier is 1.
 The above post was thanked by: Amenah (05-28-2011), bebix (05-28-2011), lollypop (05-28-2011), star123 (05-30-2011), struggle (05-28-2011)

#2
05-28-2011
 USMLE Forums Master Steps History: 1 + CK Posts: 623 Threads: 62 Thanked 289 Times in 186 Posts Reputation: 299

the ans is D
 The above post was thanked by: Claus_CU (05-28-2011), lollypop (05-28-2011)
#3
05-28-2011
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,357 Threads: 194 Thanked 3,268 Times in 881 Posts Reputation: 3278

d) the probability that each parent is a carrier is 1
 The above post was thanked by: Claus_CU (05-28-2011)

#4
05-28-2011
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This one is easy, Its D.
 The above post was thanked by: Claus_CU (05-28-2011), lollypop (05-28-2011)
#5
05-28-2011
 USMLE Forums Addict Steps History: 1 + CS Posts: 113 Threads: 8 Thanked 89 Times in 61 Posts Reputation: 99

Quote:
 Originally Posted by eesfee This one is easy, Its D.
Well, all the questions are easy, if you know how
 The above post was thanked by: bebix (05-28-2011), eesfee (05-28-2011)
#6
05-28-2011
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Why not B?
#7
05-28-2011
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,357 Threads: 194 Thanked 3,268 Times in 881 Posts Reputation: 3278

Quote:
 Originally Posted by usluipek Why not B?
answer B) "the probability that the older unaffected sister of the affected child is a carrier is 1 in 2."

Since this is an AR disease, both parent are carriers... you will have:
two Aa
one AA
one aa

Then, among the "unaffected", the probability of being a carrier is 2/3.
 The above post was thanked by: sana akbar (05-28-2011), star123 (05-30-2011), struggle (05-28-2011), usluipek (05-28-2011)
#8
05-28-2011
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Thanks, bebix,
#9
05-29-2011
 USMLE Forums Addict Steps History: 1+CK+CS+3 Posts: 109 Threads: 29 Thanked 20 Times in 13 Posts Reputation: 30

please explain me whole concept of dominant and recessive disease?
#10
05-29-2011
 USMLE Forums Master Steps History: Not yet Posts: 557 Threads: 69 Thanked 1,049 Times in 351 Posts Reputation: 1069

Quote:
 Originally Posted by amresh kumar please explain me whole concept of dominant and recessive disease?
Dominant means you only need one copy of the gene to be affected, recessive means both copies of the gene need to be mutated for someone to be affected. Some key features:
Dominant:
1. There are no carriers because if you have one copy by definition you have the disease.
2. Only 1 parent need to have the disease for the child to be at risk, and if one parent has it the probability of the child getting it is 1/2

Recessive:
1. If only one parent is a carrier or has the disease, there is 0 probability that the child will get the disease
2. If both parents are carriers or have the disease, you need to calculate probabilities based on the genotypes you are given.
 The above post was thanked by: lucas05 (07-24-2011), Mashee (07-20-2011), star123 (05-30-2011)
#11
05-29-2011
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@bebix nice and short explanation. good job.
#12
05-29-2011
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thanks buddy
#13
05-30-2011
 USMLE Forums Scout Steps History: Step 1 Only Posts: 33 Threads: 6 Thanked 9 Times in 4 Posts Reputation: 19

Quote:
 Originally Posted by dr_ram The answer should be ''B''. @bebix nice and short explanation. good job.
 The above post was thanked by: dr_ram (05-31-2011)
#14
05-30-2011
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,357 Threads: 194 Thanked 3,268 Times in 881 Posts Reputation: 3278

Quote:
 Originally Posted by moonjane11 answer is d i guess
Yes, the correct answer is D.
I think @dr_ram meant to say why can´t be B)

"Since this is an AR disease, both parent are carriers... you will have:
two Aa
one AA
one aa

Then, among the "unaffected", the probability of being a carrier is 2/3."
 The above post was thanked by: dr_ram (05-31-2011)
#15
05-31-2011
 USMLE Forums Scout Steps History: Not yet Posts: 55 Threads: 2 Thanked 93 Times in 27 Posts Reputation: 103
oops

Quote:
 Originally Posted by moonjane11 answer is d i guess
i am sorry for the disastrous error in typing. the correct answer is D. my mind was pre-occupied with something else while typing. my apologies for that.
thank you for the correction moonjane.
 The above post was thanked by: bebix (05-31-2011)

#16
06-08-2011
 USMLE Forums Addict Steps History: CK Only Posts: 178 Threads: 61 Thanked 250 Times in 105 Posts Reputation: 260

D is correct. The child is "aa" and must have received each gene from each of his parents. Since CF is AR condition and the parents are phenotypically healthy, they should be carrier for the disease, Thus they surely are "Aa".
#17
07-24-2011
 USMLE Forums Scout Steps History: Not yet Posts: 75 Threads: 33 Thanked 4 Times in 3 Posts Reputation: 14

Quote:
 Originally Posted by bebix answer B) "the probability that the older unaffected sister of the affected child is a carrier is 1 in 2." Since this is an AR disease, both parent are carriers... you will have: two Aa one AA one aa Then, among the "unaffected", the probability of being a carrier is 2/3.
I agree with u but for the probability of been a carrier is 2/3. I think the probability of been a carrier is 2/4. The "unaffectd" is just an adjective for the sister, not related to the probability ( I mean not probability of been a carrier from chances of unaffected.
Note: genetics is about chances of the individual fertilized ovum or person, not the chances of the whole family.-
That is a fertilized ovum has chances of( autosomal recessive )
1 chance to be AA= normal
2 " ". ". Aa= carrier
1. ". ". aa= disease manifest
So the probability of been normal is 1:4
" " Carrier is 2:4
" " " disease manifest 1:4
Now considering the family
If the family (God's will) all the children could b normal, all may manifest the disease, or mixture depending on thief individual outcome from thief chance

For autosomal dormant( the presence of only one gene manifest the disease)
1 chance of AA= the only normal chance
2 " " Aa= disease manifest
1 " " aa= disease manifest
Probability of having the disease is 3:4
". Been normal is 1:4
This is for both parent with gene Aa
I hope this willhelp someone

Last edited by Bibi; 07-24-2011 at 04:40 AM. Reason: typing error
 The above post was thanked by: lucas05 (07-24-2011)
#18
07-24-2011
 USMLE Forums Scout Steps History: Not yet Posts: 75 Threads: 33 Thanked 4 Times in 3 Posts Reputation: 14

Quote:
 Originally Posted by Bibi I agree with u but for the probability of been a carrier is 2/3. I think the probability of been a carrier is 2/4. The "unaffectd" is just an adjective for the sister, not related to the probability ( I mean not probability of been a carrier from chances of unaffected. Note: genetics is about chances of the individual fertilized ovum or person, not the chances of the whole family.- That is a fertilized ovum has chances of( autosomal recessive ) 1 chance to be AA= normal 2 " ". ". Aa= carrier 1. ". ". aa= disease manifest So the probability been normal is 1:4 " " Carrier is 2:4 " " " disease manifest 1:4 Now considering the family If the family (God's will) all the children could b normal, all may manifest the disease, or mixture depending on thief individual outcome from thief chance For autosomal dormant( the presence of only one gene manifest the disease) 1 chance of AA= the only normal chance 2 " " Aa= disease manifest 1 " " aa= disease manifest Probability of having the disease is 3:4 ". Been normal is 1:4 This is for both parent with gene Aa I hope this willhelp someone note
typing error earlier
#19
07-24-2011
 USMLE Forums Addict Steps History: Not yet Posts: 151 Threads: 39 Thanked 52 Times in 29 Posts Reputation: 62
D

d) the probability that each parent is a carrier is 1
#20
07-24-2011
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,357 Threads: 194 Thanked 3,268 Times in 881 Posts Reputation: 3278

Quote:
 Originally Posted by Bibi I agree with u but for the probability of been a carrier is 2/3. I think the probability of been a carrier is 2/4. The "unaffectd" is just an adjective for the sister, not related to the probability ( I mean not probability of been a carrier from chances of unaffected. Note: genetics is about chances of the individual fertilized ovum or person, not the chances of the whole family.- That is a fertilized ovum has chances of( autosomal recessive ) 1 chance to be AA= normal 2 " ". ". AA= carrier 1. ". ". AA= disease manifest So the probability been normal is 1:4 " " Carrier is 2:4 " " " disease manifest 1:4 Now considering the family If the family (God's will) all the children could b normal, all may manifest the disease, or mixture depending on thief individual outcome from thief chance For autosomal dormant( the presence of only one gene manifest the disease) 1 chance of AA= the only normal chance 2 " " Aa= disease manifest 1 " " aa= disease manifest Probability of having the disease is 3:4 ". Been normal is 1:4 This is for both parent with gene Aa I hope this willhelp someone AA= normal A note
AR disease - carrier (Aa)
If they ask among the unaffected, is 2/3 = conditional probability
If they ask "just" the probability is 1/2

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