Introduction
Watch segment 3, “One Wrong Letter”, of the PBS “Nova” program, “Cracking The Code of Life”. The first two segments of “Cracking The Code of Life” provide a good background for understanding this particular disease. Watch them if you are interested. However, these first two segments are not required. If you have trouble with viewing the video, click on “Advanced Options” which will open the full program in another window. Click to the third segment, called “One Wrong Letter,” starting at around 15:00 minutes into the program.
Disease Description
Tay-Sachs disease is an autosomal recessive disorder that results in a degeneration of the nervous system, producing symptoms at about 6 months of age. Affected children lose hearing, sight, and the ability to move. Children with Tay-Sachs disease normally die by age 3.
The gene involved in this disease is HEXA, located on chromosome #15 (8th from the bottom of the right-hand list of gene locations). The normal, dominant allele for the HEXA gene codes for the alpha subunit of the enzyme β-hexosaminidase. The HEXA enzyme is packaged in the lysosomes of nerve cells and breaks down a fatty material called GM2 ganglioside. This fatty material builds up and degrades the nerve cells of a person who has two recessive alleles for the HEXA gene locus.
Questions:
Question 1: Autosomal Recessive Trait (6 points)
In “One Wrong Letter”, Allison and Tim have discovered that their baby is suffering from Tay-Sachs disease. After some gene testing, they find that they are both heterozygous for the HEXA gene that causes the disease. Both of them have inherited one dominant allele and one recessive allele. For an autosomal recessive disease like Tay-sachs, a heterozygous person is called a “carrier” because they carry a hidden recessive allele that can be passed on to their children.
Part A: Draw a simple, but accurate diagram of the chromosomes in one of Tim’s cells in his testes at the beginning stage of meiosis. LABEL the following terms and HIGHLIGHT EACH TERM ONCE in your diagram: homologous chromosome pair #15, the two sex chromosomes, the HEXA gene locus, the dominant (HEXA) alleles, the recessive (hexa) alleles, sister chromatids.
Part B: Make a similar drawing of one of Allison’s cells in her ovary at the beginning stage of meiosis, include her #15 chromosomes and sex chromosomes. Include labels for the terms in the list above.
Question 2 (6 points)
Draw a series of three sets of diagrams that follow the two #15 chromosomes, the HEXA (dominant) and hexa (recessive) alleles, and the sex chromosomes in one parent cell undergoing meiosis to produce four sperm cells.
Diagram Set #1: Illustrate the chromosomes lined up in the center of the cell during the first division of meiosis. SHOW THE TWO POSSIBLE CHROMOSOME ALIGNMENTS of the #15 and sex chromosomes (HINT: remember independent assortment).
Diagram Set #2: Illustrate all possible combinations of the #15 and sex chromosomes in the cells at the end of the first division of meiosis based on the alignment of chromosomes in your drawings in Diagram Set #1.
Diagram Set #3: Illustrate all possible combinations of #15 and sex chromosomes that could be found in the sperm cells at the end of the second division of meiosis based on the alignment of chromosomes in your drawings in Diagram Set #2. In each of your three sets of diagrams LABEL and HIGHLIGHT the locations of the dominant and recessive alleles for the HEXA gene locus and the sex chromosomes (X or Y).
Question 3 (4 points)
What percentage of Allison’s egg cells that she produces during her lifetime will carry the recessive (hexa) allele and the X chromosome? Do not simply give a number. Support your answer with a Punnett square.
Question 4 (4 points)
If Allison and Tim have another child:
a. What is the probability (chance) that they will have a boy who is heterozygous for the Tay-Sachs gene locus?
b. Will this child have Tay-Sachs disease?
c. Show and explain how you arrived at your answer. (Punnett square with an explanation – Be sure to indicate the appropriate genetic combination(s) in your Punnett square.)
Question 5: Autosomal Dominant Trait (5 points)
Fatal Familial Insomnia (FFI) is caused by an autosomal dominant allele that produces an abnormal version of a prion protein normally present in the brain. The abnormal proteins fold unnaturally, clump together, destroy nerve cells, and cause spongy holes in the brain. People who carry just one dominant allele for FFI develop the disease in middle age. Symptoms progress fairly rapidly from mild insomnia and panic attacks, to hallucinations, to complete lack of ability to sleep, and finally, death. Average age of onset is 50 with death following 7 to 36 months later. See also, FFI in Wikipedia.
The gene associated with FFI is located on chromosome #20 (see the third gene locus from the top on left list).
A young woman learns that her mother is developing symptoms of FFI disease. The young woman’s mother is heterozygous for the FFI disease. Her father is normal (homozygous recessive).
a. Is there a chance that the young woman’s child will develop FFI disease? (The father of the child is homozygous recessive for the FFI gene.)
b. Show or explain how you arrived at your answer. (Punnett square with an explanation – Be sure to indicate the appropriate genetic combination(s) in your Punnett square.)
Question 6. Aneuploidy and Nondisjunction (10 points)
Triple X Syndrome (47 XXX) results when a person inherits two sex chromosomes from one of her parents, and a third from the other parent, resulting in a condition in which the individual has three X chromosomes. Triple X women are normal women in all regards. The extra X chromosome seems to have no real detrimental effect. There is some statistical data that indicate triple-X women are a little taller than average. Triple-X girls may be a little quieter as babies than normal girls.
See the “Abnormal chromosome numbers” section of the Lesson 10 Lecture for more information on aneuploidy and nondisjunction.
a. Draw a diagram showing how a mistake during meiosis in the development of gametes would result in a gamete with two X chromosomes.
b. Diagram or explain how this would result in a child with the XXX chromosome combination.
c. Could the mother or the father (or both) be “responsible” for this aneuploid condition in a child? Explain your choice.
Question 7: Sex-linked Recessive (X chromosome) (5 points)
Severe combined immunodeficiency (SCID), or Bubble Boy disease, is an X-linked recessive disorder. Affected individuals have little or no immune response. You may have heard about the condition as “bubble boy disease”. With no immune system, a person is susceptible to recurrent infections from many common diseases that most of us consider minor. A mutated, recessive gene fails to produce an interleukin receptor protein that is necessary for the immune system to function and protect a person from common pathogens.
A man and a woman are having a child. The woman is a carrier for SCID. The man does NOT have SCID. (Note this was updated at 8:20 pm on 4/10/16).
a. What is the probability that their child will be a boy who does not have SCID?
b. Show how you arrived at your answer. (Punnett square solution with an explanation – Be sure to indicate the appropriate genetic combination(s) in your Punnett square.)
