Download Lecture Outline 9/15 Chi-square Test for Independence Chi

Chi-square
Test for Independence
Lecture Outline 9/15
• Linkage of genes on chromosomes
• Test Cross AaBb x aabb
• Results:
– Estimating recombination fraction
• Chi-square tests for independence
– Predicting offspring phenotypes when
genes are linked
– Three-point mapping
•
•
•
•
A-BA-bb
aaBaabb
Total = 200
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Are these loci independent?
If not, How far apart are they?
Chi-square
Test for Independence
Chi-square
Test for Independence
• 2 way table of gamete frequencies for each
locus
• Test if genes assort independently into GAMETES
• Results of test cross:
•
•
•
•
A-BA-bb
aaBaabb
70
20
25
85
What did the parental gametes have to be?
70
20
25
85
A
a
total
B
70
25
95
b
20
85
105
total
90
110
200
Total = 200
Expect?
What is the
Prob( A and b),
if the loci are
independent?
Number of gametes that had big A and little b
Chi-square
Test for Independence
Chi-square
Test for Independence
• 2 way table of gamete frequencies for each
locus
Expected value =
Row total * Column total / Grand Total
B
A
a
total
70
25
95
b
20
total
90
47.25
85
105
110
200
Expected NUMBER = 0.236 * 200 = 47.25
What is the
Prob( A and b),
if the loci are
independent?
P(A) = 90/200
P(b) = 105/200
Prob (A and b) =90/200 * 105/200
= 0.236
B
b
total
A
a
total
70
25
95
E=95*90/200
E=42.75
E=95*110/200
E=52.25
20
85
E=47.25
E=57.75
90
110
easy formula
105
200
1
Chi-square
Test for Independence
Chi-square
Test for Independence
Degrees of freedom=
(Rows-1) * (Cols-1)
X2=Sum((O-E)2/E)
A
B
b
total
a
total
70
25
95
E=42.75
E=52.25
20
85
E=47.25
E=57.75
90
110
B
105
b
200
total
A
a
total
70
25
95
E=42.75
E=52.25
20
85
E=47.25
E=57.75
90
110
X2=
•
•
•
•
•
• Are these loci independent?
A-B- 70
A-bb 20
aaB- 25
aabb 85
Total = 200
– No
(s ignificant lack of fit: 60.14 >> 3.84)
• How far apart are they?
17.37 +14.21+15.71+12.85 =
105
200
60.14
Compare this to the
table of X2 for 1 df
Predicting phenotypes in
crosses with linked genes
Testcross AaBb x aabb
• Results:
<-- This formula is the same as before
• In maize, the genes g and r are 20 cM apart on
Chromosome 10. What are the expected phenotypic
classes in the cross:
– what fraction of the gametes
are recombinant?
gr/GR X gr/gr
• What would you expect if the map distance was zero?
• What would you expect if they were unlinked?
So,
• What if r=0.20?
45/200 = 0.225
= 22.5 cM
Predicting phenotypes in
crosses with linked genes
Three-point crosses
Cross:
Possible
Offspring
Genotypes
Gamete type
GgRr
Ggrr
ggRr
Ggrr
Parental
Recombinant
Recombinant
Parental
Map distance = 20 cM
Gray body,Red-eye,normal wing
x
yellow body, white eye, miniature wing
(+++ x ywm)
Total
recombinants
= 20%
Shorthand notation for phenotypes
Results:
+++
++m
+w+
+wm
y++
y+m
yw+
ywm
758
401
1
16
12
0
317
700
--What is the order of
these genes on the X
chromosome?
--What are the map
distances?
2
Three-point crosses
1. Find reciprocal pairs
Results:
+++
++m
+w+
+wm
y++
y+m
yw+
ywm
758
401
1
16
12
0
317
700
Three-point crosses
3. Compare parentals and double
recombinants to find the middle gene
Results:
2. Which are the
parentals?
single recombinants?
double recombinants?
+++
++m
+w+
+wm
y++
y+m
yw+
ywm
758
401
1
16
12
0
317
700
+
+
+
+
y
w
+
+
m
y
w
m
See fig 4-13
Three-point crosses
4. Calculate the recombination fraction
between neighboring pairs of loci
Results:
+++
++m
+w+
+wm
y++
y+m
yw+
ywm
758
401
1
16
12
0
317
700
• Find reciprocal pairs
• Use double recombinants to find gene
order
– Which are the double recombinants?
y
?
w
m
Recombination fraction between y and w
=single crossovers + double crossovers
=(16+12+1+0)/2205
=1.3%
Now you try it:
dan
da+
d+n
d++
+++
++n
+a+
+an
Three point crosses
299
5
131
65
308
8
124
60
Total = 1000
1. Which are the parental classes?
2. Which are the single
recombinants?
3. Which are the double
recombinants?
4. What is the gene order?
• Then estimate recombination between
the two pairs of loci to get map
distances
– Don’t forget to add in the double crossovers!
Interference
• You can predict the frequecy of double crossovers:
– it should be the product of the two single crossovers
• R(ab and bc) = R(ab) * R(bc)
Why is it a product?
• Often the observed number is less than expected due
to interference
(rewrite in correct gene order if necessary)
5. What are the recombination
distances for each neighboring
pair of genes?
Coincidence coefficient
= observed double crossovers / expected double crossovers
• Interference can differ among organisms
– In Drosophila, there is complete interference (no double crossovers) to about
10 cM;
– in yeast, only partial interference at 3 cM
– There is essentially no interference beyond 20 or 30 cm in most organisms
3