Chi Square

ACTIVITY: Chi square on Human Traits

OBJECTIVE:
Establish and compare the results on human genetic traits between the expected values and the observed values by using a Chi Square in order to determine if the observed values are due to chance or some other factor.

TIME: 40 min.

PROCESS SKILLS:
- Observe
- Analyze
- Compare/Contrast
- Synthesize

MATERIALS:
- Paper
- Pencil
- Calculator or Spreadsheet (Optional)

INVESTIGATION PROCEDURE:

  1. Several human traits are controlled by a single set of alleles. For example, some persons can bend their thumb so that it appears behind their hand; others can roll their tongue into a U shape. Some have earlobes that hang freely, and others have earlobes that are attached to the head. The expected phenotype for this trait can be found in the table below called Expected Values of Human Traits.

    PHENOTYPE TABLE OF THE EXPECTED VALUES OF HUMAN TRAITS.
    Human Traits Human Trait is present Human Trait is absent
    Flexion of Thumb 13.84 % 86.16 %
    Ear Lobe 70.76 % Free 29.24 % Attached
    Tongue Roll 74.58 % 25.42 %25.42 %

    Note: These results for the phenotype of human traits are for the population of Monterrey Mexico. The results are from the IMSS Research Center.

  2. Collection the results of the whole class for the different types of human traits.
    Traits Flexion of Thumb NO Flexion of Thumb Attached Ear Lobe Free Ear Lobe Tongue Roll NO Tongue Roll
    Number of students 9 6 8 7 13 2


  3. Before doing the calculations for Chi Square determine the number of individuals in class. Use this formula:

    population of class X the percent of the expected value = the expected value

    Example: There are 36 students in the class. The expected value for flexionof thumb = 36 students X 13.84/100% = 4.9824 rounded to 5

    Calculations for Chi Square of a given human trait.
    A. Calculation for Chi Square of flexion of thumb
    Phenotype Observed number*
    O    		 Expected number*
    E    		 Deviation =
    (O-E)    		 Deviation squared(O-E)2 (O-E)2 = X2
    -----
    E
    Flexion of Tumb
    No flexion of Tumb
    Totals Sum(O-E)2
    ------- =
    E

    Total X2 =

    B. Calculation for Chi Square of ear lobe
    Phenotype Observed number
    O    		 Expected number
    E    		 Deviation =
    (O-E)    		 Deviation squared(O-E)2 (O-E)2 = X2
    -----
    E
    Attached Ear Lobe
    Free Ear Lobe
    Totals Sum(O-E)2
    ------- =
    E

    Total X2 =
    C. Calculation for Chi Square of tongue roll
    Phenotype Observed number
    O    		 Expected number
    E    		 Deviation =
    (O-E)    		 Deviation squared(O-E)2 (O-E)2 = X2
    -----
    E
    Tongue Roll
    No Tongue Roll
    Totals Sum(O-E)2
    ------- =
    E

    Total X2 =


    Interpretation of Results:

    Using the Chi square table, interpret the Chi Square value you have just obtained.

    Step 1: Find n value Remember that for each human trait there are only two classes of data. Therefore the n value for your Chi Square table will be 2-1= 1. n is always 1 for each Chi Square of the human traits that you considered.

    Step 2: Find the probability number. Trace your calculated chi square number for one human trait along the row which is n=1. When you have located your chi square number, then look for the corresponding number on the first top row which begins with letter P.99.

    Step 3: Accept or reject observed values for a human trait If your probability is 0.05 or less than 0.05, then observed values for a human trait are rejected. If your probability is greater than 0.05, then observed values for a human trait are accepted.

    Step 4: Repeat Steps 1 to Steps 3 for chi square value of each human trait .

    Note: If the sample size is small (like 36 students in a class), then it would be very difficult to accept the observed results. A larger sample size like 300 students would give better observed results.

    Conclusion
    Did your observed results agree with the expected results? If not, offer an explanation for this lack of agreement.

    HOMEWORK:
    In PEGS there are 266 males and 191 females who are 17 yr. old. In PEGL there are 103 males and 82 females who are 17 yr. old.. You must establish the size of your sample and determine the observed value for your chosen trait of that sample. Finally calculate the Chi Square for this chosen trait. How does this Chi Square value for large sample size compares to the Chi Square value for a smaller sample of the same trait?

    The excercise above was from ITESM in Mexico