Download Chi Squared Analysis

GENETIC
DATA
ANALYSIS
SUM RULE- The combined
probability of two events that are
mutually exclusive is the sum (add!)
of the individual probabilities.
Clue: look for “or”
Q: What’s the probability of rolling
a ‘five’ or a ‘six’ on one sixsided die?
A: 1/6 + 1/6 = 1/3
GENETIC EXAMPLE: MONOHYBRID CROSS
What is the probability that the F2
offspring has the dominant phenotype
(is either GG or Gg)?
P:
F1:
GG x gg =
Gg
Gg x Gg =
F2: 1/4 GG: 1/2 Gg: 1/4 gg
3 out of 4 dominant
1. MONOHYBRID COMPLETE DOMINANCE RATIO
P: Homozygous Cross
BB x BB =
4 BB : 0 Bb : 0 bb
bb x bb=
0BB : 0Bb : 4bb
F2: Heterozygous Cross
Bb x Bb=
1 BB : 3Bb: 1bb
3 dominant : 1 recessive
3:1 RATIO
2. DIHYBRID COMPLETE DOMINANCE RATIO
Heterozygous= 9 : 3 : 3 : 1 Ratio
9 dominant in both loci B
 3 dominant in first locus, recessive in second locus
 3 dominant in second locus, recessive in first locus
 1 recessive in both loci

9+3+3+1= 16 (possible offspring genotypes)
9 Smooth, Yellow (SY)
3 wrinkle, Yellow (sY)
3 Smooth, green (Sy)
1 wrinkle, green (sy)
3. TRIHYBRID, TETRAHYBRID, ETC

Doing a punnet square would be excessive, so
use the ratio!
Monohybrid
3:1
Dihybrid
9:3:3:1
Trihybrid
27:9:9:9:3:3:3:1
3/4 B
1/4 b
3/4 × 3/4 = 9/16 BD
3/4 × 1/4 = 3/16 Bd
1/4 × 3/4 = 3/16 bD
1/4 × 1/4 = 1/16 bd
3/4 × 3/4 × 3/4 = 27/64 BDR
3/4 × 3/4 × 1/4 = 9/64 BDr
3/4 × 1/4 × 3/4 = 9/64 BdR
1/4 × 3/4 × 3/4 = 9/64 bDR
3/4 × 1/4 × 1/4 = 3/64 Bdr
1/4 × 3/4 × 1/4 = 3/64 bDr
1/4 × 1/4 × 3/4 = 3/64 bdR
1/4 × 1/4 × 1/4 = 1/64 bdr
PRODUCT RULE-The probability
of several independent events is
the product (multiply!) of the
individual probabilities. Clue: look
for “and”
Q: You roll two dice. What’s the
probability of getting a ‘2’ on the first
one and a ‘5’ on the second one.
A: 1/6 x 1/6 = 1/36
GENETIC EXAMPLE:
P: AA bb CC DD ee ff x aa BB cc dd EE FF
F1: Aa Bb Cc Dd Ee Ff x Aa Bb Cc Dd Ee Ff
Q: What proportion of F2 progeny will be
AA bb Cc DD ee Ff ?
A: 1/4 * 1/4 * 1/2 * 1/4 * 1/4 * 1/2 = 1/1024
CHI SQUARED
ANALYSIS
CHI SQUARED ANALYSIS
 Determines
if data is “good”
FORMULA: X2 = Σ (O – E)2
E
O: observed values
E: expected values
CHI SQUARED NULL HYPOTHESES
1.
2.
Null Hypothesis: there is no significant difference
between the expected and observed result

If accepted, you are claiming that mating is random
and so is segregation and independent assortment

P Value is greater than 0.05
Reject Null Hypothesis:
•
you are seeing that the deviation between observed
and expected is very far apart…something nonrandom must be occurring….
•
P Value is less than 0.05
DEGREES OF FREEDOM


Must determine the degrees of freedom
Degrees of freedom: # of phenotypic
possibilities in your cross minus 1
LET’S LOOK AT A FRUIT FLY CROSS
x
P: Black body,
eyeless
F1: all wild
wild
F1 x F1
5610
1896
1881
622
RESULTS ANALYSIS
1. Determine the cross that you were using
Dihybrid Complete Dominance
2. What is the expected outcome of this cross?




9/16 wild type (yellow body, eyes)
3/16 normal body eyeless (yellow body, no eyes)
3/16 black body wild eyes (black body, no eyes)
1/16 black body eyeless (black body, eyes)
3. CONDUCT THE ANALYSIS: CREATE A PERFECT, “EXPECTED” 9:3:3:1 RATIO
A. Divide total individuals by 16
10009/16 = 626
B. Multiply by perfect ratio#
626*9= 5,634 for 2 domani loci (wild type)
Phenotype
Observed
Expected
Wild
5,610
5,634
(=626*9)
Eyeless
1,881
1,878
(=626*3)
Black Body
1,896
1,878
(=626*3)
Eyeless, Black Body
622
TOTAL
10,009
626
(=626*1)
4. Apply the Chi Squared formula to each phenotype:
(5610 - 5630)2/ 5630 = 0.07
(1881 - 1877)2/ 1877 = 0.01
(1896 - 1877 )2/ 1877 = 0.20
(622 - 626) 2/ 626 = 0.02
 2= .30
5. Determine Degrees of Freedom = # of phenotypes – 1
4-1= 3
6. Find the value closest to chi squared on Chi Squared
Distribution Table, ex. 0.35
CHI-SQUARE DISTRIBUTION TABLE
Reject Null
Hypothesis
Accept Null Hypothesis
Probability (p)
Degrees of
Freedom
0.95
0.90
0.80
0.70
0.50
0.30
0.20
0.10
0.05
0.01
0.001
1
0.004
0.02
0.06
0.15
0.46
1.07
1.64
2.71
3.84
6.64
10.83
2
0.10
0.21
0.45
0.71
1.39
2.41
3.22
4.60
5.99
9.21
13.82
3
0.35
0.58
1.01
1.42
2.37
3.66
4.64
6.25
7.82
11.34
16.27
4
0.71
1.06
1.65
2.20
3.36
4.88
5.99
7.78
9.49
13.38
18.47
5
1.14
1.61
2.34
3.00
4.35
6.06
7.29
9.24
11.07
15.09
20.52
6
1.63
2.20
3.07
3.83
5.35
7.23
8.56
10.64
12.59
16.81
22.46
7
2.17
2.83
3.82
4.67
6.35
8.38
9.80
12.02
14.07
18.48
24.32
8
2.73
3.49
4.59
5.53
7.34
9.52
11.03
13.36
15.51
20.09
26.12
9
3.32
4.17
5.38
6.39
8.34
10.66
12.24
14.68
16.92
21.67
27.88
10
3.94
4.86
6.18
7.27
9.34
11.78
13.44
15.99
18.31
23.21
29.59
6. CONCLUDE CHI SQUARED ANALYSIS

With
degrees of freedom, my chi
square value is
, which gives me a p value of
%, I therefore
my null
hypothesis
“With 3 degrees of freedom, my chi squared value is
0.35, which gives a P value of 95%, therefore I
accept my null hypothesis. “
This means that 95% of the time when our observed data is this
close to our expected data, this deviation is due to random
chance.
• P value is greater than 0.05, therefore accept our null hypothesis
•
EXCEPTIONS TO
MENDELIAN
RATIOS
EXCEPTIONS TO MENDELIAN RATIO
1.
2.
3.
4.
5.
Lethal Alleles
Epistasis
Unusual sex linkage
Sex influenced inheritance
Genetic Anticipation
1. LETHAL ALLELES: MANX CATS
LETHAL ALLELES
F1:
F1:
F2:
Mm
1 MM
x
Mm
2 Mm
1 Lethal: 2 Manx:
1 mm
1 Normal
F2 phenotypic ratio is 2:1 instead of 3:1
OTHER LETHAL MUTATIONS

Achondroplasia (humans)- short limbed dwarfism

Yellow body color (domestic mice)

Curly wings (Drosophila flies)
2. EPISTASIS



Genetic interaction between two (or more) loci
One gene modifies the phenotypic effects of
another gene
Normal dihybrid ratio is altered from 9:3:3:1 to 9:3:4
9 Wild
 3 Black *novel/new genotype
 4 Albino

Agouti:
wild type
P:
F1:
BB CC
agouti
x
Bb Cc
agouti
bb cc
albino
Simple dominant phenotype?
F2: 9/16 B- C- 3/16 bb C- 3/16 B- cc 1/16 bb cc
albino
agouti
albino
black
F2 Phen. ratio: 9 agouti : 3 black : 4 albino
novel phenotype
Epistasis
Normal dihybrid
ratio is altered
from 9:3:3:1 to
9:3:4
C and B gene
have an epistatic
interaction
PRACTICE PROBLEM

In Labrador retrievers, coat color is controlled by two loci each
with two alleles B,b and E,e respectively. When pure breeding
Black labs with genotype BB EE are crossed with pure breeding
yellow labs of genotype bb ee the resulting F1 offspring are
black. F1 offspring are crossed (Bb Ee x Bb Ee). Puppies appear
in the ratio:
9/16 black;

3/16 chocolate;
4/16=1/4 yellow.
What genotypes correspond to these three phenotypes?
9/16 B- EB- E-
3/16 B- ee
B- ee
3/16 bb E_ 1/16 eebb
bb E- and bb ee
3. UNUSUAL SEX LINKAGE
4. ENVIRONMENT DEPENDENT ALLELES
PHENOTYPES ARE NOT ALWAYS A DIRECT
REFLECTION OF GENOTYPES
1.
2.
Temperature-sensitive alleles- Siamese color
pattern
Nutritional effects: phenylketonuria
SIAMESE
OR
“HIMALAYAN”
•Different allele of the C locus that causes lighter color
is Temperature sensitive.
•Cooler the body area, the darker the color
5. GENETIC ANTICIPATION
Causes later generations in a family to be more
severely affected by a disease
 Increases the # of triplet repeats in the fragile
area of the X chromosome through the
generations

Ex.
 Huntington disease
 Fragile-X syndrome
 Kennedy disease
 Myotonic muscular dystrophy
*Huntingtons
Disease is a
Dominant
Disorder