Part ATay-Sachs disease is an autosomal recessive disorder, so, o.pdf

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ankitmobileshop235

Part A: Tay-Sachs disease is an autosomal recessive disorder, so, only a homozygous recessive person will have the disease. 1 in 300 individuals of neither descent will be carrier. A child can have the disease, only if both his/her parents are carriers. Let the genotype of Tay Sachs disease be \'tt\'. So, carrier will be \'Tt\' According to HW equilibrium, 2pq= 1/300 Given data is p = 0.0167 q = 0.983 (This data is totally wrong). Please note that \'q\' is the recessive Tay-Sachs allele and \'p\' is the dominant one. So, the frequency of recessive allele cannot be more than the dominant one. When two carriers cross, 25% chance is there of having the genotype \'tt\'. So, chances of having a child with Tay-Sachs disease = 25% of the total children born or 25% of (1/300 * 1/300) Part B. If one individual is of Ashkenazi descent, then chances of being a carrier is 1/30 If the other parent is normal, then chances of being a carrier is 1/300 When two carriers cross, chances of having homozygous recessive genotype = 25% So, 25% chance of having an affected child is there i.e 25% of (1/30 * 1/300) Part C: Again 25% chance of having an affected child is there. You can also write as: 25% of (1/30 * 1/30) Part D: Again 25% chance of having an affected child is there. You can also write as: 25% of (1/30 * 1/30) Solution Part A: Tay-Sachs disease is an autosomal recessive disorder, so, only a homozygous recessive person will have the disease. 1 in 300 individuals of neither descent will be carrier. A child can have the disease, only if both his/her parents are carriers. Let the genotype of Tay Sachs disease be \'tt\'. So, carrier will be \'Tt\' According to HW equilibrium, 2pq= 1/300 Given data is p = 0.0167 q = 0.983 (This data is totally wrong). Please note that \'q\' is the recessive Tay-Sachs allele and \'p\' is the dominant one. So, the frequency of recessive allele cannot be more than the dominant one. When two carriers cross, 25% chance is there of having the genotype \'tt\'. So, chances of having a child with Tay-Sachs disease = 25% of the total children born or 25% of (1/300 * 1/300) Part B. If one individual is of Ashkenazi descent, then chances of being a carrier is 1/30 If the other parent is normal, then chances of being a carrier is 1/300 When two carriers cross, chances of having homozygous recessive genotype = 25% So, 25% chance of having an affected child is there i.e 25% of (1/30 * 1/300) Part C: Again 25% chance of having an affected child is there. You can also write as: 25% of (1/30 * 1/30) Part D: Again 25% chance of having an affected child is there. You can also write as: 25% of (1/30 * 1/30).

Part A:
Tay-Sachs disease is an autosomal recessive disorder, so, only a homozygous recessive person
will have the disease. 1 in 300 individuals of neither descent will be carrier. A child can have the
disease, only if both his/her parents are carriers.
Let the genotype of Tay Sachs disease be 'tt'. So, carrier will be 'Tt'
According to HW equilibrium, 2pq= 1/300
Given data is p = 0.0167 q = 0.983 (This data is totally wrong). Please note that 'q' is the
recessive Tay-Sachs allele and 'p' is the dominant one. So, the frequency of recessive allele
cannot be more than the dominant one.
When two carriers cross, 25% chance is there of having the genotype 'tt'.
So, chances of having a child with Tay-Sachs disease = 25% of the total children born or 25% of
(1/300 * 1/300)
Part B.
If one individual is of Ashkenazi descent, then chances of being a carrier is 1/30
If the other parent is normal, then chances of being a carrier is 1/300
When two carriers cross, chances of having homozygous recessive genotype = 25%
So, 25% chance of having an affected child is there i.e
25% of (1/30 * 1/300)
Part C:
Again 25% chance of having an affected child is there.
You can also write as: 25% of (1/30 * 1/30)
Part D:
Again 25% chance of having an affected child is there.
You can also write as: 25% of (1/30 * 1/30)
Solution
Part A:
Tay-Sachs disease is an autosomal recessive disorder, so, only a homozygous recessive person
will have the disease. 1 in 300 individuals of neither descent will be carrier. A child can have the
disease, only if both his/her parents are carriers.
Let the genotype of Tay Sachs disease be 'tt'. So, carrier will be 'Tt'
According to HW equilibrium, 2pq= 1/300
Given data is p = 0.0167 q = 0.983 (This data is totally wrong). Please note that 'q' is the
recessive Tay-Sachs allele and 'p' is the dominant one. So, the frequency of recessive allele

cannot be more than the dominant one.
When two carriers cross, 25% chance is there of having the genotype 'tt'.
So, chances of having a child with Tay-Sachs disease = 25% of the total children born or 25% of
(1/300 * 1/300)
Part B.
If one individual is of Ashkenazi descent, then chances of being a carrier is 1/30
If the other parent is normal, then chances of being a carrier is 1/300
When two carriers cross, chances of having homozygous recessive genotype = 25%
So, 25% chance of having an affected child is there i.e
25% of (1/30 * 1/300)
Part C:
Again 25% chance of having an affected child is there.
You can also write as: 25% of (1/30 * 1/30)
Part D:
Again 25% chance of having an affected child is there.
You can also write as: 25% of (1/30 * 1/30)

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2. cannot be more than the dominant one. When two carriers cross, 25% chance is there of having the genotype 'tt'. So, chances of having a child with Tay-Sachs disease = 25% of the total children born or 25% of (1/300 * 1/300) Part B. If one individual is of Ashkenazi descent, then chances of being a carrier is 1/30 If the other parent is normal, then chances of being a carrier is 1/300 When two carriers cross, chances of having homozygous recessive genotype = 25% So, 25% chance of having an affected child is there i.e 25% of (1/30 * 1/300) Part C: Again 25% chance of having an affected child is there. You can also write as: 25% of (1/30 * 1/30) Part D: Again 25% chance of having an affected child is there. You can also write as: 25% of (1/30 * 1/30)