Question

tay-sachs disease is an inborn errors of metabolism that results in death, often by the age of 2. You are a genetics counselor interviewing a phenotypically normal couple who tell you the male had a female first cousin(on the father's side) who died of 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 mating have been between related individuals. Assume that the trait is very rare. Draw a pedigree of the families of those couple, showing relevant individuals. Calculate the probability that both the male and female are carries of TSD.

Answer 1

Answer:

Explanation:

Deterioration of the nerve and brain cells are caused by clusters of an enzyme called gangliosides. Hexosaminidase is an enzymes that reduces the build up of this gangliosides. An individual that lacks this Hexosaminidase is prone to suffer from Tay-sachs disease. This  Tay-sachs disease is an example of recessive autosomal disorder.

Now, this type of disease will be  autosomal disorder if the offspring has both defective alleles. So the offspring can be 1/4 probability of being affected  aor be a carrier (i.e 2/4 probability) . In short, it is possible to be an a carrier which is heterozygous in nature. Though the heterozygous individual doesn't have symptoms of the disease but it will be expressed in the following generation.

(a)

The objective is to draw  a pedigree of the families of those couple, showing relevant individuals from the information given.

SEE BELOW FOR THE ATTACHED FILE  OF THE PEDIGREE CHART

(b)  Calculate the probability that both the male and female are carries of TSD.

For the female:

Since the mother is unaffected; she should be a carrier (Tt) ; as such, the probability of being a carrier is   $\dfrac{2}{3}$ . However, the female father is not a carrier; from the  assumption given in the question  that the traits are rare; we assume that the female father is homozygous dominant (TT)

Thus ; the cross between the female father and mother will be :

      T                 T

T     TT               TT

t       Tt               Tt

we will see from the above punnet square that the probability of being a carrier is 1/2 Tt and 1/2 TT  normal.

Now the probability of the female being a carrier will be :

P(mother) × P (female) = $\dfrac{2}{3} *\dfrac{1}{2} = \dfrac{2}{6} \\ \\ = \dfrac{1}{3}$

For the male;

since the father is unaffected; he should be a carrier (Tt) ; as such the probability of being a carrier is $\dfrac{1}{2}$. From the information given; the male's mother is said not be a carrier, Hence, the mother is homozygous dominant TT.

However, the cross between the male's mother TT and the father will be:

      T        T

T    TT       TT

t      Tt          Tt

we will see from the above punnet square that the probability of being a carrier is 1/2 Tt and 1/2 TT  normal.

Thus; the probability of the female being a carrier will be ;

P(father) × P(male)  $=\dfrac{1}{2} *\dfrac{1}{2} \\ \\ = \dfrac{1}{4}$

Using the product rule to determine the probability that both male and female are carriers of Tay-Sachs diseases (TSD) is:

P(male) × P (female)  $=\dfrac{1}{3} *\dfrac{1}{4}$

$= \dfrac{1}{12}$

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