[struggle]

A young couple presents to ur office seeking prenatal counseling. The husband suffers from cystic fibrosis and is very concerned that his child will get the disease. The family pedigree is diagrammed below..
((its difficult for me to draw a pedigree.. but husband and wife both are cousins and their parents didn't manifest the disease but one of his wife's sister got the disease..))
what is the chance that this child will get the disease?
A. 1/16
B. 1/8
C. 1/4
D. 1/3
E. 2/3
F. 3/4

 

[FrontalBossing]

From the father the kid is for sure getting the CF gene, so that 100% chance, which is equal to 1. The mother is not affected, so that means there is a 2/3 chance she has the CF gene (she's either a carrier or Homozygous Normal) and then there is the a 50% shot of getting that affected CF gene from the mom if she indeed does have it so that's 1/2.

1 x (2/3) x (1/2) = 1/3 chance of getting the disease.

 

[Mondoshawan]

C. 1/4

I would answer C. 1/4

One of the wife's alleles is normal. Since neither of her parents manifested the disease but both were carriers, then there is a 50% chance that her other allele is the CF allele. Multiply that by the 50% chance that she will pass it on, as FrontalBossing said, and that makes 25%.

 

[aktorque]

I would answer C. 1/4

One of the wife's alleles is normal. Since neither of her parents manifested the disease but both were carriers, then there is a 50% chance that her other allele is the CF allele. Multiply that by the 50% chance that she will pass it on, as FrontalBossing said, and that makes 25%.

Excellent...Mondoshawan

simply you say that the mom being a carrier for CF is 1/2 (50%)
Chance that she passes it on to her child is 1/2 (50%)

now you multiply this two together 1/2 * 1/2 = 1/4 (25%) chance that this child will get the disease.

 

[apx85]

I agree with Frontal Bossing.

The parents of the mother have to be carrier's since her sister has the disease but they do now show the phenotype (CF is auto recessive)

Mothers parents are both Aa

Therefore, the mother's potential phenotypes are: 1/4 AA, 2/4 Aa, or 1/4 aa but since she does not have CF, she cannot be aa, leaving 3 possibilities. She has a 1/3 chance to be AA and 2/3 to be Aa based on probabilities.

The mother has a 50% chance of passing on the a allele if she is a carrier, so overall, her chance of passing on the a (recessive) allele is 2/3 x 1/2 = 1/3

The father has 100% chance of passing on CF gene, therefore the answer should be 1/3

Hope that made sense/is correct :sorry:

 

[doctorsmonsters]

Chances of the mother being a carrier are 2/3.
See, both her parents are carriers. 1/4 will be homozygous, 2/4 hetero and 1/4 homozygous normal. Since the mother doesn't have the disease, she is one of the three remaining possible combination so chances of being a carrier are 2/3.

The father has the disease, he's definitely going to transfer the diseased gene. If the mother is a carrier (2/3 probability) then the probability that she will transfer the gene is 1/2, so we have to independent probabilities, and we'll multiply them, we get 1/3, which is the answer.

 

[doctorsmonsters]

haha, just realized after posting, while typing that, apx had explained that already.)

 

[Mondoshawan]

Oh - yeah, I see what you're saying. That makes sense... maybe I was thinking about it backwards

 

[sairam]

since the father is affected and once of his wife's sister was affected the answer can be
1.the ratio of children that will manifest with this pathology will be 1/4
2,the ratio of children who will be jus carriers without any manifestations will be 2/4..'

 

[struggle]

1/3 is the answer..
Since cystic fibrsis is an autosomal recessive disease and males always transmit the disease so chances of transmitting this disease to offspring is 100% or probability is 1..
mother's parents didnt manifest the disease but one of her sibling got the disease so her chances of getting the carrier gene is 2/3..(as in autosomal diseases if mother and father both are carrier, 25% of offsprings will manifest the disease,50% will be carrier and 25% will be homozygous normal..so as one of her sibling got the disease,out of the remaining 3 sibling,2 will be carrier,,so probability is 2/3))..
the chance that she will transmit the carrier gene to her child is 50% or 1/2..
so multiplying all..
1*2/3*1/2 =1/3..

 

[laurier]

plz can someone give me a link or any sources where i can understand well this genetic topic, and how to get these probabilities.
i'm really confused, and can't answer genetic questions like this?

thank's in advance

 

[struggle]

u can listen to kaplan genetics video..it helped me..

 

[laurier]

unfortunatly i dont have kaplan video :toosad:.
i will be gratfull if someone share a copy of them with me.

thank's in advance