Please use the "Talk to Ed and Mike" forum in Moodle if you wish to discuss the assignment further.

Completion of these questions will help you achieve most of the objectives for lectures #11, #12, and #13

Question #1. Autosomal Recessive Trait (5 points)

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 loose 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.

A couple are interested in having a child, but have learned that the man is a carrier for the disease and the woman is homozygous dominant. What is the chance that their first child will be born a carrier of Tay-Sachs disease?

Show how you arrived at your answer.

Watch segment 3 of the PBS "Nova" program, "Cracking The Code of Life" - "One Wrong Letter". This is an interesting case study of a couple dealing with Tay-Sachs disease

Answer to Question #1

The man is a carrier (heterozygous) for this autosomal recessive disease (HEXA, hexa).

The woman is also a carrier (HEXA, HEXA).

Meiosis will produce sperm that are either HEXA or hexa.

Meiosis will produce eggs that are only HEXA.

The genotypes of the possible children are represented by the red and blue letters in the boxes of the Punnett square.

There is a 50% chance the first child will be a carrier for this autosomal recessive disease (heterozygous hexa, HEXA).

Homozygous dominant children will be healthy and not be able to pass the recessive gene to their childeren.

Question #2. Autosomal Dominant Trait (5 points)

Huntington disease is caused by an autosomal dominant mutant allele that produces an abnormally long version of a protein normally present in the brain. The result is a degeneration of nerve function causing loss of motor control and personality changes, ultimately leading to death. The symptoms do not become apparent until a person is in his or her 40s. See also Huntington's Disease Society of America

The gene associated with Huntington disease is located on chromosome #4. The protein that is produced by the gene is "huntingtin". The mutation adds repeated triplets of nucleotides to the gene and produces a longer than normal protein with enhanced, abnormal function. The mutant protein produced by the mutated dominant allele overshadows the function of the normal protein produced by the recessive allele.

A teenage mother learns that her father is developing symptoms of Huntington disease. Her father is heterozygous. Her mother is normal (homozygous recessive).  Is there a chance that the young woman's child has inherited the dominanat Huntington disease gene? (The father of the child is homozygous recessive for the Huntington gene.)

Show how you arrived at your answer.

Answer to Question #2

The teenage mother does not know if she has the dominant allele that causes Huntington disease, an autosomal dominant disorder.  Her father has the disease and is heterozygous, so there is a 50% chance she could inherit the #4 chromosome with the dominant allele from her father.  The teenage mother's mother is homozygous recessive and free of Huntington's disease. The teenage mother can only get the recessive allele from her mother.  We are left with the conclusion that there is a 50% chance the teenage mother is heterozygous and will develop the disease and a 50% chance she is heterozygous and disease free.

IF the teenage mother is heterozygous, she and her husband who is homozygous recessive could produce children like this:

A child of this union would have a 50% chance of inheriting the dominant, disease causing allele.

IF the teenage mother is homozygous recessive, she and her husband who is also homozygous recessive could produce children like this:

A child of this union would have no chance of inheriting the dominant, disease causing allele.

In this case, there is a one in two, or 50% chance the teenage mother is heterozygous (H, h) and will develop Huntington disease.  IF the mother is hpmozygous recessive her child has a 0% chance of being heterozygous.  IF the mother is heterozygous her child has a 50% chance of being heterozygous.  In this case there IS a chance the child will inherit Huntingtondisease.  The chance of the child inheriting Huntington disease is 50% X 50% = 25%.

Question #3. Aneuploidy and Nondisjunction (5 points)

XYY Syndrome (47 XYY) results when a person inherits two sex chromosomes from one of his parents, and a third from the other parent, resulting in a condition in which the individual has three sex chromosomes (an X and 2 Y chromosomes). The affected individual develops as a male. Individuals with this condition generally are of taller stature and may have acne. Some boys with XYY syndrome are physically more active and may have delayed mental maturation.

More information about XYY Syndrome:

See the "Abnormal chromosome numbers" section of lecture outline #12 for more information on aneuploidy and nondisjunction.

Draw a set of diagrams showing how a mistake during meiosis in the development of gametes would result in a gamete with two Y chromosomes.

Show how this would result in a child with the XYY chromosome combination.

Would the mother or the father be "responsible" for this aneuploid condition in a child? Explain your choice.

Answer to Question #3

Extra or missing chromosomes can occur in a gamete (egg or sperm) as the result of a mistake during meiosis called nondisjunction. Nondisjunction can occur at two times during meiosis. Homologous chromosomes may fail to separate from each other and travel together to one of the two cells during the first division of Meiosis as illustrated in this movie from from Hironao NUMABE, M.D.

The production of egg cells with two Y chromosomes can occur by nondisjunction ONLY in Meiosis II
See also Hoefnagels, pg. 185, fig. 9.11).  In the illustration in this link from your text, mentally label the two chromosomes in the first cell on the left as X and Y and then follow them through meiosis to see how a sperm can get two Y chromosomes.  The 4 cells on the far right illustrate the zygotes after fertilization with an egg carrying one X chromosome (as well as the other 22 chromosomes not illustrated).

Nondisjunction could happen in meiosis I during the production of sperm, HOWEVER this mistake would have produced sperm with two both X and Y chromosomes and two sperm with no sex chromosomes.

When a sperm carrying two Y chromosomes fertilizes an egg cell carrying one X chromosome, the resulting offspring is a male who has two Y chromosomes and one X chromosome. The term that applies is aneuploidy, in this case noted as 47 XYY.

In this problem, only dad could be the "culprit" because mom has no Y chromosome.

Remember, of course, that "dad" had no conscious role in producing a defective sperm. It is just a matter of chance.

Question #4. Two Gene Genetics and X-linkage (5 points)

Autosomal Recesive

Fish odor syndrome is a human autosomal recessive condition that results in body odor that resembles rotting fish.

The gene responsible for this genetic disorder is located on chromosome #1 (Zoom in on lower portion of the yellow section below the centromere - nine gene loci below the yellow section on the right-hand list of gene loci.

The fishy odor of people with the "disease" is caused by body emissions of the chemical, trimethylamine (TMA). TMA is a digestion byproduct of foods high in choline, found in fish, eggs, and liver. The dominant allele of the "fish odor" gene produces a protein, which acts as an enzyme that breaks down TMA. The name of the enzyme is flavin-containing mono-oxygenase 3 (FMO3). The recessive allele results from a mutation of this gene that produces a non-functional FMO3 enzyme.

Sex-linked Recesive (X chromosome)

Severe combined immunodeficiency (SCID) 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.

Bubble Boy disease from MSNBC

A man and a woman are having a child.

The man is a carrier (heterozygous) for the fish odor syndrome and does NOT have SCID.

The woman is also a carrier for the fish odor syndrome. The woman's father has SCID disease. The woman's mother is homozygous dominant for the SCID gene locus.

What is the probability that their child will be a girl who has the fish odor syndrome AND SCID? Show how you arrived at your answer.

Answer to Question #4

The man is a carrier, heterozygous for fish odor syndrome (FMO3, fmo3), an autosomal recessive trait. He does not have SCID, a sex linked (X chromosome) recessive trait. This tells us that he has a dominant allele (SCID) on his X chromosome, and of course, no corresponding gene locus for SCID on his Y chromosome.

As a result of independent assortment during meiosis, the man can produce these types of sperm in equal proportions:

We know that the woman is a carrier, heterozygous (FMO3, fmo3) for fish odor syndrome so half her eggs will have the dominant allele and half will have the recessive allele for that trait.

We can figure out from her parents' genetotypes and phenotypes that the woman is heterozygous for SCID (SCID-X came from her mother and  scid-X from her father), so half her eggs will contain the the dominant allele for SCID (SCID) on one X chromosome and half will contain recessive allele for SCID (scid) on her other X chromosome.

The woman's eggs will have these genotypes in equal proportions.

FMO3, SCID-X

FMO3, scid-X

fmo3, SCID-X

fmo3, scid-X

Here is one way we can figure out how their children might turn out:

		The Woman's Eggs
		FMO3, SCID-X	FMO3, scid-X	fmo3, SCID-X	fmo3, scid-X
	FMO3, SCID-X	FMO3, FMO3, SCID-X, SCID-X	FMO3, FMO3, SCID-X, scid-X	FMO3, fmo3, SCID-X, SCID-X	FMO3, fmo3, SCID-X, scid-X
The Man's	FMO3, Y	FMO3, FMO3, Y, SCID-X	FMO3, FMO3, Y, scid-X	FMO3, fmo3, Y, SCID-X	FMO3, fmo3, Y, scid-X
Sperm	fmo3, SCID-X	fmo3, FMO3, SCID-X, SCID-X	fmo3, FMO3, SCID-X, scid-X	fmo3, fmo3, SCID-X, SCID-X	fmo3, fmo3, SCID-X, scid-X
	fmo3, Y	fmo3, FMO3, Y, SCID-X	fmo3, FMO3, Y, scid	fmo3, fmo3, Y, SCID-X	fmo3, fmo3, Y, scid-X

The answer to the question is that the probability that their first child will be a girl who has fish odor syndrome AND SCID Disease (fmo3, fmo3; scid-X, scid-X) is 0%.

You can also figure this out by looking at each characteristic individually and multiplying the probabilities.

The probability that two people heterozygous for fish odor syndrome will produce a homozygous recessive child is 25%. (FMO3, fmo3 crossed with FMO3, fmo3) - do the 4-square Punnett square if you need to.

The probability that these two people will produce a girl child who has SCID is 0%. (SCID-X, Y crossed with SCID-X, scid-X) - do the 4 square Punnett square if you need to.  All the girls produced by this union will get the dominant SCID allele on the X chromosome form the father.  Half the girls will be carriers of SCID and half will be homozygous dominant.  None of the possible girls will be homozygous recessive.

Multiply the probabilities for the two traits.

25% chance of fmo3, fmo3 multiplied by 0% chance of scid-X, scid-X gives you 0% chance of fmo3, fmo3; scid-X, scid-X.

Note that the SCID gene locus is sex-linked on the X chromosome, so gender and SCID are NOT independent when calculating probabilities.

The way that most people go wrong in a question like this is in forgetting that the Y chromosome has no gene locus for SCID disease as well as a whole lot of other gene loci found on the X chromosome. Males have a greater chance of inheriting X-linked recessive traits because they have no corresponding gene locus on their Y chromosome to counteract a recessive X-linked allele inherited from a carrier mother.