Focus Activity 2.  Vocabulary Crossword   tip

Focus Activity 3.  Sex Determination & Syndromes:  The Test   tip

Activity #1--15 points total possible

Background information -- Students have this on their page -- you may want to present this first in lecture format with "chalk talk".  Notice "Lyon Hypothesis" in yellow.

Exactly how different are men and women? Each has 46 chromosomes in every one of their cells.  Each cell has two of each of 22 alleles and one pair of sex chromosomes. Both males and females have at least one X chromosome on which such important genes as glucose - 6 - phosphate dehydrogenase resides. The other sex chromosome is either another full X chromosome containing another full complement of genes or a much smaller Y chromosome with essentially no genes which are allelic to the X chromosome. Much of the Y consists of redundant multi-copy non-expressing genetic material with a few sexual characteristic genes thrown in for good measure.

If all the genes on the X chromosome were concerned with sex this arrangement might be
acceptable. However, it is apparent that most X chromosome genes have nothing whatsoever to do with sexuality but instead are functional and common in both sexes. Do females then have twice as much of enzymes such as glucose - 6 - dehydrogenase? The answer is no because one of the X chromosomes in females in inactivated a mechanism known as “dosage compensation”.

In 1949 Murray L. Barr and E.G. Bertram described a small darkly staining body that was present in the somatic cells of female cats. Further studies found this dark staining body to be found in the somatic (body) cells of both human and feline females. It was subsequently named the “Barr Body” and could be observed most easily in cells scraped from the lining of the mouth (buccal cells). It was found to be an X chromosome which was inactivated and not actively transcribing.  It was observed in all cells except the sex cells. (Presumably it is necessary for two active X chromosomes for proper oocyte development.)

In 1962, Mary Lyon proposed an explanation for this phenomenon. According to her hypothesis, in the process of early development (about the 32 cell stage) an X chromosome becomes randomly inactivated and stays inactivated for the rest of the cell’s life. In addition, any daughter cells derived from that original cell also maintain the X as inactivated. This X chromosome is called “heterochromatin” as opposed to the term “euchromatin” which refers to the chromosome on which actively transcribing genes are located. As to which of the two X alleles is inactivated it appears that it is a completely random process which happens by chance alone. In some cells the inactive X may be paternal (from the father) in origin and in some cells it may be maternal (from the mother) in origin.

Since each female is made of a mixture of cells with different active X chromosomes then every female is in reality a mosaic for all the genes which are actively being expressed on the X chromosomes. Different alleles will in all probability reside on different X chromosomes.
Evidence that the condensed, presumably inactive, X was truly inactive for gene expression was first found in mice heterozygous for the dominant and recessive coat color alleles, wild type (brown) and albino (white). On cytogenetic analysis it was discovered that, although the gene for mouse color is located on an autosome other than the X chromosome a mutation had occurred. A portion of the chromosome with the mouse color gene on it had broken and reattached (translocated) to one of the X chromosomes.

Therefore when the X chromosome with the mouse color gene was inactivated, the autosomal
gene was inactivated as well. Since the inactivated gene was the dominant gene in those cells it was the recessive trait which began to be expressed. In cells where the normal X was inactivated the dominant gene was able to express.  Since inactivation did not occur until about the 32 cell stage the mouse was mosaic for color producing a coat some of which was albino and some of which was the wild type brown color.  This phenomenon produced a strain referred to as the “Tortoise” color.  In this activity you will explore the probability of any particular cell and the progeny of that cell carrying the active dominant gene.

1. What color will the resulting mouse be if the embryo is Male?
2. How do you know this?

This particular mouse type experiences random X inactivation at the 16 cell stage. At this stage of embryo division either the normal X will inactivate and become unable to express its genes or the abnormal X will inactivate.

3. If the abnormal X inactivates what color fur will the clone of cells from this original cell be?
4. If the normal X inactivates what color fur will the clone of cells from this original cell be?

The mice on the following page are divided into 16 sections. These represent the 16 cells of the trophoblast that were present when inactivation occurred. (We are assuming the inactivation occurred at 16 cells for the simplicity of the activity - it probably occurs a little later.) Since inactivation of any specific X chromosome is a random process we will simulate this with the flip of a coin. Determine the gene which is being expressed in each of the 16 original cells which will be come clones of cells within the mouse and give it a mottled appearance by flipping the coin. If heads then the normal X is inactivated. If tails then the abnormal X is inactivated.

Teacher’s Answer Key:
1. Brown
2. The X with the gene for color will always be active since there is only one X chromosome in a male mouse.
3. White
4. Brown
5. One out of 65536 combinations—(1/2)¹⁶
6. One out of 65536 combinations—(1/2)¹⁶
7. Males = 0 tortoise, 1/2 brown, 1/2 albino
Females = 1/2 tortoise, 0 brown, 1/2 albino
8. There would not be any cells in the female that have a gene inactivated that is critical for
survival. Any cell that had a critical gene inactivated would die and only those that had
normal active critical genes would survive in the embryonic tissue. Therefore the adult mouse
would be made only of cells with the normal X inactivated.
Note: The mouse coloring activity will depend on the values obtained by flipping the coin. There is no right answer but the mouse coloration must agree with the coin flipping chart.

Teacher's Notes

It is best to have students study known cells with Barr bodies first so they can most easily identify them. It is difficult to discern Barr bodies without a little practice and without a fairly decent microscope. The best way to do this is to prepare some sample buccal smears prior to the class and then choose the best of these to use for practice and demonstration slides. A magnification of at least 400 is suggested. The reason that Barr bodies are not found in every cell in women is simply due to the limitations of a two dimensional microscope. The Barr body
may easily be present but because it is on the "back side" of the cell it can not be viewed. Point out to students that the reason 11-19% is considered a positive is because when cytogenetics were doing their quality control studies they found that on the average a normal female will show a percentage of 11-19% of cells with a
"viewable" Barr body with a confidence of 95 percent.

Reagents:

     Cresyl violet acetate (substitute for Cresyl echt violet): Matheson Coleman and Bell 10 g powder lots.  Certified by the Biological Stain Commission.

     Cresyl violet acetate working solution (1!): 1 g cresyl violet acetate in 100 mL 50% ethanol. Prepare fresh every 3 months.

     Ethanol, 95% pure grain: Mallinchrodt Chemical Works. 1 gal lots.

     Ethanol, 50%: 45 mL deionized water and 50 mL 95% ethanol.

     Fixative: 10 mL glacial acetic acid and 30 mL methanol.

     Glacial acetic acid: E.I. duPont de Nemours and Company (Inc.): Reagent grade, meets ACS specifications. 5 lb lots.

     Methanol: Mallinchrodt Chemical Works. Methanol anhydrous (absolute) acetone free. Analytical reagent. 1 pt lots.

     Xylol: Mallinchrodt. 1 pt bottles.

Preparation before the Class:

     Prior to the laboratory, prepare the following:
        1.Working cresyl violet acetate solution 
        2.Fixative 
        3.It is easiest if on the second day of laboratory the teacher sets up a series of jars of stain, alcohol and xylol in a hood for easier and more efficient staining. Xylol should always be kept in a hood so that breathing of the fumes can be kept to an absolute minimum. 
        4.It is most efficient if on the first day of this laboratory, depending on the length of class, the teacher remove the slides from the fixative and allow to air dry in a designated area. 
        5.Labeling is critical. Xylol and alcohol will remove most types of marking pen. Therefore it is advisable to put paper labels (stickers) on the slides at the beginning of the staining and buccal cell collection.
 

Teacher's Answer Key 

   1.The X chromosome

   2.One

   3.Zero

   4.Two

   5.One

   6.Duchenne Muscular Dystrophy, Hemophilia, Color Blindness, etc...

   7.Possibly. If the female were heterozygous and the "normal" X is inactive then the abnormal X would be expressed in some cells.

   8.Yes. It would give the sex of the baby. This would help to determine the probability of having expression of an abnormal X chromosome gene.

   9.A doctor would do a buccal smear in cases of ambiguous genitalia, suspected sex chromosome abnormalities or any time when chromosomal sex needs to be quickly obtained.

  10.We are looking at a three dimensional cell with a two dimensional instrument. Some Barr bodies may be on the backside or are out of view.

  11.Since there are many genes on the X chromosome which are active in every human regardless of sex, the inactivation provided for dosage compensation. In this manner both male and female have equal amounts of the products of these genes.

  12.A fixative is a mixture of acetic acid and methanol. It helps to stabilize the cell structure while making them stick to the microscope slide.

  13.- 20. Fill in the following chart.