Three point mapping: A, B, and C genes (alleles A and a, B and b, C...
Three point mapping:
A, B, and C genes (alleles A and a, B and b, C and c) with complete dominance for all.
The testcross is: AaBbCc x aabbcc
The progeny are seen below.
ABC: 75 aBC: 75
ABc: 5 aBc: 345
AbC: 345 abC: 5
Abc: 75 abc: 75
1) If the genes are not linked, could this happen by chance alone? Write your assumptions. do the chi-square/show the math, and do you accept or reject the hypothesis that the genes are not linked and assort independently of each other?
2) Assume the genes are linked. Establish the Gene Order and map distances for the 3 loci, and draw a final map.
Homework Answers
Answer:
1).
|
Phenotype |
Observed(O) |
Expected (E) |
O-E |
(O-E)2 |
(O-E)2/E |
|
ABC |
75 |
125 |
-50 |
2500.00 |
20.00 |
|
ABc |
5 |
125 |
-120 |
14400.00 |
115.20 |
|
AbC |
345 |
125 |
220 |
48400.00 |
387.20 |
|
Abc |
75 |
125 |
-50 |
2500.00 |
20.00 |
|
aBC |
75 |
125 |
-50 |
2500.00 |
20.00 |
|
aBc |
345 |
125 |
220 |
48400.00 |
387.20 |
|
abC |
5 |
125 |
-120 |
14400.00 |
115.20 |
|
abc |
75 |
125 |
-50 |
2500.00 |
20.00 |
|
Total |
1000 |
1000 |
1084.80 |
Chi-square value = 1084.80
Degrees of freedom = number of phenotypes – 1
Df = 8-1 = 7
The chi-square value of 1084.80 is greater than the critical value of 14.07 (at 5% p value). Hence, we reject null hypothesis and the genes are not assorted independently. The three genes are linked.
2).
Answer:
Hint: Non-recombinant (parental) combinations are produced more in number than recombinants. The parental (non-recombinant) genotypes is AbC / aBc
1).
If single crossover occurs between a & b
Normal combination: Ab/aB
After crossover: AB/ab
AB progeny=75+5=80
ab progeny = 5+75=80
Total this progeny = 160
Total progeny = 1000
The recombination frequency between a&b = (number of recombinants/Total progeny) 100
RF = (160/1000)100 = 16%
2).
If single crossover occurs between b&c
Normal combination: bC/Bc
After crossover: bc/BC
bc progeny= 75+75=150
BC progeny = 75+75=150
Total this progeny = 300
The recombination frequency between b&c = (number of recombinants/Total progeny) 100
RF = (300/1000)100 = 30%
3).
If single crossover occurs between a & c
Normal combination: AC/ac
After crossover: Ac/aC
Ac progeny=5+75=80
aC progeny = 5+75=80
Total this progeny = 160
Total progeny = 1000
The recombination frequency between a&c = (number of recombinants/Total progeny) 100
RF = (160/1000)100 = 16%
Recombination frequency (%) = Distance between the genes (mu)
b--------16mu------a----16mu----c
The order of genes = bac
Fingal gene map = b--------16mu------a----16mu----c
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