Chisquare test in genetics

1

• Chi-square test in genetics
• testing if a given genetic data set fit a
  certain model

For example, a dihybrid testcross should give a
1111 ratio between all four categories of
phenotypes. But the actual data obtained through
experiments will be different from the predicted
ratio. What will be the probability of getting
these discrepancies by chance?
P Prob discrepancy btw actual vs expected
numbers due to chance If P gt 0.05, discrepancy
is blamed on chance alone If P lt 0.05 something
else is affecting the observed discrepancy
2

• How to use the Chi-square test in genetics
• two values need to be calculated

The degree of freedom (df) No. of category
minus 1 or df 4 -1 in the case of dihybrid
testcross. Chi-square value ? observed no -
expected no.2/ expected no.
? means combine all categories values
expected no means the predicted ratios based on
the Mendels Laws
Example 1 AaBb x aabb
dihybrid cross Use branching diagram method
A gene B gene
1/2 Bb
------- 1/2 Aa

1/2 bb -------- ---------gt
1/2 Bb
-------- 1/2 aa

1/2 bb --------
3

• How to use the Chi-square test in genetics
• two values need to be calculated

The degree of freedom (df) No. of category
minus 1 or df 4 -1 in the case of dihybrid
testcross. Chi-square value ? observed no -
expected no.2/ expected no.
? means combine all categories values
expected no means the predicted ratios based on
the Mendels Laws
Example 1 AaBb x aabb
dihybrid cross Use branching diagram method
A gene B gene
1/2 Bb
------- 1/4 AaBb
1/2 Aa
1/2 bb -------- 1/4
Aabb ---------gt
1/2 Bb -------- 1/4
aaBb 1/2 aa
1/2 bb
-------- 1/4 aabb
4

• How to use the Chi-square test in genetics
  (continue)
• actual calculation

Phenotype ratio Observed No Expected No O -
E (O-E)2 (O-E)2/E AaBb 1/4 17 1/4
x5012.5 Aabb 1/4 8 1/4 x5012.5
aaBb 1/4 11 1/4 x5012.5 Aabb 1/4 14 1
/4 x5012.5 Total 1 50 50 0
Phenotype ratio Observed No Expected No O -
E (O-E)2 (O-E)2/E AaBb 1/4 17 1/4
x5012.5 4.5 20.25 1.62 Aabb 1/4 8 1/4 x5012.5
-4.5 20.25 1.62 aaBb 1/4 11 1/4 x5012.5
-1.5 2.25 0.18 Aabb 1/4 14 1/4
x5012.5 1.5 2.25 0.18 Total 1 50 50 0 3.60
df (degree of freedom) 4-1 3 Chi-sqare value
3.60
5
Chi square table of critical values
df (degree of freedom) 4-1 3 Chi-sqare
value (?2) 3.60
P Prob discrepancy btw actual vs expected
numbers due to chance If P gt 0.05, discrepancy
is blamed on chance alone (thus cannot reject the
initial model) If P lt 0.05 something else is
affecting the observed discrepancy
6
Chi square table of critical values
The greater ?2 values ---gt the smallest p values
means something else is causing the deviation
between observed vs expected numbers!
7
More examples of the chi square tests
Draw sample 12 times out of the Mendel box
containing 40 tubes with a ratio of 1111
phenotypes (Exp.1) vs 5115 phenotypes (Exp.2)
Vial (V) vs Tube (v) ---- shape Taped (T)
vs Untaped (t) ---- color
Vt
8

• initial model for Exp. 1(these two traits are
  independently sorted,
• and thus will follow the
  1111 ratio)

Phenotype ratio Observed No Expected No O -
E (O-E)2 (O-E)2/E VT 1/4 1/4
x503 Vt 1/4 1/4 x503 vT 1/4 1/4
x503 vt 1/4 1/4 x503 Total 1 12 12
0
9

• initial model for Exp. 2 (these two traits are
  independently sorted,
• and thus will follow the
  1111 ratio)

Phenotype ratio Observed No Expected No O -
E (O-E)2 (O-E)2/E VT 1/4 1/4
x503 Vt 1/4 1/4 x503 vT 1/4 1/4
x503 vt 1/4 1/4 x503 Total 1 12 12
0
10

• Summary of Mendels Laws

• Modifications (or extensions) to these Mendels
  conclusions (Introduction to other lectures)
• alleles without clear dominance/recessiveness
  ------ incomplete dominance and codominance
• alleles with lethal effects
• - alleles with more than two alleles ----
  multi-allele traits (human ABO blood types)
• - alleles at different genes interact in
  controlling one phenotypic trait --- Epistasis
  (Golden retriever)
• - genes located in sex chromosomes (human
  color-blindness and hemophilia)
• two different genes are located on the same
  chromosome linkage and genetic mapping
• the two genes will not be sorted
  independently due to the chromosomal linkage
• deviation from the 1111 ratio in
  testcross.