Laboratory 9.  Transmission genetics and the chi-square goodness of fit test

    Each of us carries two copies of each of our genes (exceptions: Y chromosome and mitochondria).  We got one from our mother and the other from our father.  Sometimes the two copies are identical; we then say we are homozygous for that gene.  Alternatively, the two copies could be different from each other; we then say we are heterozygous for that gene.  If we become a parent we will give only one copy to each offspring.  When sperm or eggs are made, the two copies of the genes segregate and only one of them is included.  For any gene in which we are homozygous it doesn't matter which copy we give; they are both the same.  For any gene in which we are heterozygous, our offspring will have a 50% chance of receiving one version and a 50% chance of receiving the other version.  This is the basis for Gregor Mendel's principle of segregation.  It's the same as throwing a coin for heads and tails. 
    In this week's lab we will examine the concept of chance in relation to gene segregation.  We will learn to use the chi-square goodness of fit test.  For this lab you have permission to print out the whole lab instead of writing a pre-lab and all you have to do is answer the questions. 

Materials:
    Ears of corn
    Marker pen
Bring a calculator (simple calculator is good).

Procedure:
1.  Evaluation of color transmission.  Work in pairs:  You will have an ear of maize that is segregating for two traits: dark versus light colored kernels and also for sugary (wrinkled) versus normal kernels.  First focus on color only; we will look at the sugary trait later.  The dark colored allele is C and the colorless allele is c (homozygotes are yellow).  The ear was produced by the self pollination of an individual that was heterozygous Cc.  The expectations for color are as follows:

Expectations
(proportions)
C
(paternal)
c
(paternal)

C
(maternal) 		1/4
C/C
dark
 1/4
C/c
dark

c
(maternal) 		1/4
c/C
dark
 1/4
c/c
yellow

What is the expected proportion of dark kernels? __________

What is the expected proportion of yellow kernels? __________

    Shell off about (not exactly) one hundred kernels.   Count the dark and light kernels, ignoring whether they are wrinkled or not. 
    Fill out the following chart.  The expected number of dark kernels is 3/4 x the total counted.  The number of light kernels is 1/4 x total.  The expected numbers might not be integers. 

Chi-square
 Observed
 Expected
 (Obs - Exp)2 ÷ Exp

Dark




Light



Totals



Chi-square = Total of (Obs - Exp)2 ÷ Exp = ____________

Degrees of freedom: _______

Estimated probability of fit by chance (P): __________ (use chart below). 

Was your P value less than 0.05?  _________   Less than 0.01?  _________

2.  Evaluation of sugary transmission.  Work in pairs:  Using the same kernels, count normal (smooth) versus sugary (wrinkled like a raisin), regardless of whether the kernels are dark or yellow.  The kernels were produced by the self pollination of an individual that was heterozygous Su su.  The expectations for color are as follows:

Expectations
(proportions)
Su
(paternal)
su
(paternal)

Su
(maternal) 		1/4
Su/Su
smooth
 1/4
Su/su
smooth

su
(maternal) 		1/4
su/Su
smooth
 1/4
su/su
sugary

What is the expected proportion of smooth kernels? __________

What is the expected proportion of sugary kernels?  __________

    Fill out the following chart.  The expected number of normal kernels is 3/4 x the total counted.  The number of sugary kernels is 1/4 x total.  Again, the expected numbers may not be integers. 

Chi-square
 Observed
 Expected
 (Obs - Exp)2 ÷ Exp

Smooth




Sugary




Totals



Chi-square = Total of (Obs - Exp)2 ÷ Exp = ____________

Degrees of freedom: _____

Estimated probability of fit by chance (P): __________

Was your P value less than 0.05?  _________   Less than 0.01?  _________


3.  Evaluation of color and sugary transmission.   Work in pairs:  In the previous experiments, we considered only one trait at a time, as if the kernels were F2 progeny of a monohybrid.  Clearly, these kernels are F2 progeny of a dihybrid. 

Expectations
(proportions)
 C Su
 C su
 c Su
 c su
C Su
 1/16
C C Su Su
 1/16
C C Su su 		1/16
C c Su Su		 1/16
C c Su su
C su 1/16
C C su Su 		1/16
C C su su 		1/16
C c su Su		 1/16
C c su su
c Su
 1/16
c C Su Su		 1/16
c C Su su		 1/16
c c Su Su		 1/16
c c Su su
c su
 1/16
c C su Su		 1/16
c C su su		 1/16
c c su Su		 1/16
c c su su


What is the expected proportion of dark smooth kernels (C _ Su _)?     __________

What is the expected proportion of dark sugary kernels (C _ su su)?     __________

What is the expected proportion of yellow smooth kernels (c c  Su _)?  __________

What is the expected proportion of yellow sugary kernels (c c  su su)?  __________


    Fill out the following chart.  Again, the expected numbers may not be integers. 

Chi-square
 Observed
 Expected
 (Obs - Exp)2 ÷ Exp
Dark smooth


Yellow smooth


Dark sugary


Yellow sugary


Totals



Chi-square = Total of (Obs - Exp)2 ÷ Exp = ____________

Degrees of freedom: ________   (careful!) 

Estimated probability of fit by chance (P): __________

Was your P value less than 0.05?  _________   Less than 0.01?  _________








Chi-Square Chart. 

Chi-square chart