Tay-Sachs disease (TSD) is an inborn error of metabolism that results in death,

Question

Tay-Sachs disease (TSD) is an inborn error of metabolism that results in death, often by the age of 2. You are a genetic counselor interviewing a phenotypically normal couple who tell you the male had a female first cousin (on his father’s side) who died from TSD and the female had a maternal uncle with TSD. There are no other known cases in either of the families, and none of the matings have been between related individuals. Assume that this trait is very rare.

(a) Draw a pedigree of the families of this couple, showing the relevant individuals. (b) Calculate the probability that both the male and female are carriers for TSD.
(c) What is the probability that neither of them is a carrier?
(d) What is the probability that one of them is a carrier and the other is not?

[Hint: The p values in (b), (c), and (d) should equal 1.]

Explanation / Answer

Tay -sachs disease also called hexosaminidase A deficiency is a autosomal recessive disorder .two tay-sachs allele should be there to exhibhit disease . carriers of the single allele donot exhibhit .when both the parents are carriers , there is chance of 25% giving risk to affected child . both sexes are equally affected .

from the above question we get that ;

maternal side - her uncle is having the disease so , her parents are heterozygous for the disease .so mother will be having the 1/3 chance of having the disease, when we cross between two parents of mother that is Tt *Tt = we get Tt will be 1/3 chance of having heterozygous mother or if we cross between the homozygous parents TT* TT = we get TT 1/3 chance of disease .

so to conclude mother will be 1/3 chance of homozygous or heterozygous for the disease {TT or Tt}

paternal side - his cousin is having , in order to get a disease the person should be a carrier for the disease .considering fathers parents mother is can be TT homozygous , fathers father will be heterozygous Tt   

so if we get cross TT*Tt =offsprings will be 1/4 TT ,1/2 will be Tt,

father of being heterozygous Tt is 1/3 chance .

so to conclude ; Calculate the probability that both the male and female are carriers for TSD.

mother will be Tt act as a carrier

father will be Tt act as carrier

Tt*Tt =offsprings will be TT [ 1/4 ] ,Tt [1/2] ,tt [1/4 ].

2/36 childern will have the carrier allele Tt or 1/36 will have dominant allele TT not acting as carrier for disease

(c) What is the probability that neither of them is a carrier?

if parents not carrier they must be homozygous for disease TT *TT[2/3chance from the above caluculations for each parents ] so offsprings will be 2/3*2/3 = 4/9 =16/36 will be homozygous for the allele ,they will not exhibhit disease .

(d) What is the probability that one of them is a carrier and the other is not?

there will be two probability , TT{mother } *Tt [father] , Tt [mother} *TT [father}

TT{mother } 1/3 *Tt [father] 2/3 ;

offsprings 1/2 TT and 1/2 will be Tt ,so we get 1/2*2/3*1/2 = 2/18 = 4/36 will have chance of being Tt or 4/36 will have chance of being TT.

TT{mother } *Tt [father]

same as above there will be 4/36 will have TT or 4/36 will have Tt allele .

so to conclude

chance of child being TT

1/36+ 16/36 +4/36 + 4/36 = 25/36 will be TT

chance of being Tt

2/36 + 4/36+ 4/36 = 10/36 will have Tt

chance of being tt =1/36

25/36+10/36+ 1/36 = 36/36

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.