Part 2 - Diagnosis
A week after the appointment was made, Anna takes Jared in for the check up. The doctor first asks if Anna had taken a prenatal screening or got a chorionic villus sampling (CVS) between the 10th and 12th weeks of pregnancy. She didn't know what a CVS even was and her doctor explained to her that it is a process in which a small sample of the placenta is drawn into a needle or a small tube and the cells are analyzed for deformities. The doctor also asked if she had an amniocentesis to test for Tay-Sachs during her 15th and 18th weeks of pregnancy. She replied with "No" once again. Both Anna and her husband did not consult with a genetic counselor due to the fact there had been no severe health issues in either of the families' history. The doctor explained to her that he believed that Jared may have a recessive autosomal disease called Tay-Sachs disease. He explained that it is a disease that is caused by a mutation in the gene (HEXA) hexosaminidase A which is found on chromosome 15. It codes for an enzyme where its purpose is to eliminate fatty protein and other dangerous material that can interfere with growth, especially in the brain. The doctor examined Jared's eyes and noticed a red spot on the retina, which could be a sign of the disorder. In order to be sure the doctor drew Jared's blood and took it for testing. When the test comes back and the blood shows that Jared is lacking the enzyme, it would mean the he was, indeed, a Tay-Sachs baby.
Part 2 - Questions
Question 1. What is a karyotype and what is it used for? What would a karyotype look like from a person suffering from Tay-Sachs disease?
A karyotype is a particular set of chromosomes that an individual possesses. It is used to determine if a fetus has any abnormalities such as trisomy (the presence of an extra chromosome) or monosomy (the absence of a chromosome). The karyotype of an affected child with the disease would look normal, since the disease is characterized as a faulty gene, not an extra or missing chromosome.
Question 2. What is autosomal inheritance? Describe what is an autosome and how many are present in the human body?
Autosomal inheritance refers to dominant and recessive traits that are coded for by genes on autosomes. An autosome is any chromosome other than the sex chromosomes, and there are 22 pairs present in the human body.
A karyotype is a particular set of chromosomes that an individual possesses. It is used to determine if a fetus has any abnormalities such as trisomy (the presence of an extra chromosome) or monosomy (the absence of a chromosome). The karyotype of an affected child with the disease would look normal, since the disease is characterized as a faulty gene, not an extra or missing chromosome.
Question 2. What is autosomal inheritance? Describe what is an autosome and how many are present in the human body?
Autosomal inheritance refers to dominant and recessive traits that are coded for by genes on autosomes. An autosome is any chromosome other than the sex chromosomes, and there are 22 pairs present in the human body.
Question 3. Refer to the pedigree above. In order for an individual to have Tay-Sachs disease, a child would have to inherit two recessive alleles for the gene, one from each parent.
Part a) If Male E is affected with the disease, what is the genotype of that individual and it homozygous or heterozygous?
His genotype is aa and he is homozygous recessive.
Part b) Male D does not have Tay-Sachs disease; however, he has a son that does. What are the possible genotypes of Male D and Female C? Why are they unaffected by the disease?
Both Male D and Female C have the genotype Aa because their son would have to receive two recessive copies in order to inherit the disease. These individuals are unaffected because they are heterozygous. Only individuals who are homozygous recessive are affected with the disease.
Part c) What is the possible genotype of Male F and why?
Male F's genotype would have to be aa because he is inherited the disease. Since it is an autosomal recessive disorder, he would need two copies of the recessive gene in order to express disease.
Question 4. Explain the difference between sex-linked inheritance and autosomal inheritance. What is the mode of inheritance for Tay-Sachs disease?
Sex-linked inheritance refers to the dominant and recessive genes that are found on the sex chromosomes (pair 23). This results in one gender affected more commonly than the other gender. This results in one gender that is affected more than the other. Autosomal inheritance refers to the dominant and recessive traits that are coded for by the genes that are present on the autosomes (the chromosome pairs 1-22). Since the affected gene that characterizes Tay-Sachs disease is found on one of the autosomes (chromosome 15), it is autosomal inheritance. This means that it affects men and women equally.
Question 5. A child with Tay-sachs disease is homozygous recessive (bb). Explain the difference between heterozygous and homozygous. Predict the genotype and phenotype of the children of a heterozygous dominant couple and a homozygous couple. Use a Punnett square to support your predictions.
Homozygous is when an individual has two identical alleles for a trait, it can be either dominant (BB) or recessive (bb). Heterozygous is when an individual has two different alleles for a trait, it is represented by Bb.
Part a) If Male E is affected with the disease, what is the genotype of that individual and it homozygous or heterozygous?
His genotype is aa and he is homozygous recessive.
Part b) Male D does not have Tay-Sachs disease; however, he has a son that does. What are the possible genotypes of Male D and Female C? Why are they unaffected by the disease?
Both Male D and Female C have the genotype Aa because their son would have to receive two recessive copies in order to inherit the disease. These individuals are unaffected because they are heterozygous. Only individuals who are homozygous recessive are affected with the disease.
Part c) What is the possible genotype of Male F and why?
Male F's genotype would have to be aa because he is inherited the disease. Since it is an autosomal recessive disorder, he would need two copies of the recessive gene in order to express disease.
Question 4. Explain the difference between sex-linked inheritance and autosomal inheritance. What is the mode of inheritance for Tay-Sachs disease?
Sex-linked inheritance refers to the dominant and recessive genes that are found on the sex chromosomes (pair 23). This results in one gender affected more commonly than the other gender. This results in one gender that is affected more than the other. Autosomal inheritance refers to the dominant and recessive traits that are coded for by the genes that are present on the autosomes (the chromosome pairs 1-22). Since the affected gene that characterizes Tay-Sachs disease is found on one of the autosomes (chromosome 15), it is autosomal inheritance. This means that it affects men and women equally.
Question 5. A child with Tay-sachs disease is homozygous recessive (bb). Explain the difference between heterozygous and homozygous. Predict the genotype and phenotype of the children of a heterozygous dominant couple and a homozygous couple. Use a Punnett square to support your predictions.
Homozygous is when an individual has two identical alleles for a trait, it can be either dominant (BB) or recessive (bb). Heterozygous is when an individual has two different alleles for a trait, it is represented by Bb.
The F1 generation of the heterozygous parents has a genotype probability of 50% Bb, 25% BB and 25% bb. This couple would have a 25% chance of having a child with Tay-Sachs disease. The F1 generation of the homozygous parents would be 100% BB. This couple would have a 0% chance of having a child with the disease.
Question 6. Use the punnet square to answer the following:
Part a) What is the percentage of a child not getting the disease?
Normal child is Tt or TT. Therefore 3/4 = 75%
Part b) What is the percentage of a child carrying the disease?
Carried Child (Tt) = 2/4 = 50%
Part c) What is the percentage of having an affected child?
1/4 = 25%
Part a) What is the percentage of a child not getting the disease?
Normal child is Tt or TT. Therefore 3/4 = 75%
Part b) What is the percentage of a child carrying the disease?
Carried Child (Tt) = 2/4 = 50%
Part c) What is the percentage of having an affected child?
1/4 = 25%