Tay-Sachs disease is a hereditary disease that kills 1 in 360,000 individuals in the general population. Individuals homozygous for the defective allele have Tay-Sachs disease and die at an early age.Studies show that heterozygous individuals have a higher rate of survival against tuberculosis than the rest of the population.Biochemical tests can determine if parents are carriers.
a. What type of inheritance is demonstrated by Tay-Sachs disease?
My work:The type of inheritance that is demonstrated by Tay-Sachs disease is Autosomal Dominant Inheritance.
b. A young couple decide to have genetic screening done to determine if they were carriers of Tay-Sachs. If both individuals were carriers what percentage of their offspring would be predicted to have protection from tuberculosis, but not have Tay-Sachs disease.(Show all your work)
My work: 35%
a. How can it be dominant, if there are carriers who do not have the disease?
b. No, use Punnett Square:
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To determine the percentage of their offspring that would be predicted to have protection from tuberculosis, but not have Tay-Sachs disease, we need to understand the inheritance patterns and probabilities involved.
Tay-Sachs disease is inherited in an autosomal recessive manner. This means that it requires both parents to be carriers of the defective gene in order for their offspring to be affected with the disease. Carriers are individuals who have one normal allele and one defective allele of the gene.
We know that both individuals in the young couple are carriers of Tay-Sachs disease. Let's use the Punnett square method to determine the probabilities of their offspring having protection from tuberculosis but not having Tay-Sachs disease.
The parents are both carriers, so their genotypes can be represented as follows:
Parent 1: Tt (carrier)
Parent 2: Tt (carrier)
Using a Punnett square, we can cross their genotypes to determine the possible combinations in their offspring:
t t
T | TT 50% | Tt 50% |
t |
------------
T | TT 50% | Tt 50% |
t |
From the Punnett square, we can determine the genotypes and phenotypes of their offspring:
- 25% of the offspring will have the genotype TT, meaning they will not be carriers of Tay-Sachs disease and will also not have protection against tuberculosis. (TT - no TB protection, no Tay-Sachs disease)
- 50% of the offspring will have the genotype Tt, meaning they will not have Tay-Sachs disease but will have protection against tuberculosis. (Tt - TB protection, no Tay-Sachs disease)
- 25% of the offspring will have the genotype tt, meaning they will be affected with Tay-Sachs disease. (tt - Tay-Sachs disease)
Since we're interested in the percentage of offspring with protection against tuberculosis but not Tay-Sachs disease, we add the first two percentages together:
25% + 50% = 75%
Therefore, 75% of their offspring would be predicted to have protection from tuberculosis, but not have Tay-Sachs disease.
In conclusion, if both individuals in the couple are carriers of Tay-Sachs disease, there is a 75% chance that their offspring will have protection from tuberculosis but not have Tay-Sachs disease.