PCB 3063-Sec 799 Homework # 1 Question 6: (2 points) Consider three X-linked rec
ID: 206836 • Letter: P
Question
PCB 3063-Sec 799 Homework # 1 Question 6: (2 points) Consider three X-linked recessive Drosophila genes, a, b and c. A triple heterozygote female was crossed to a male that is mutant for all three traits. The first 1,000 offspring isolated had the following phenotypes: 462 27 16 458 23 14 What phenotypes are missing in the offspring and why? What is the order of these genes on the X-chromosome? Construct a linkage map that includes the distance between the genes. Calculate interference. a. b. e. d.Explanation / Answer
Answer:
a). Missing phenotypes are abc & +++. Due to very close linkage between the genes the double crossover is not possible. Hence, the phenotypes are not produced.
b). Order of genes on X-chromosomes is a+b/+c+
c). a--------5 cM-----c---------3 cM--------------b
d). Interference=1
Explanation:
Hint: Always recombinant genotypes are smaller than the non-recombinant genotypes.
Hence, the parental (non-recombinant) genotypes is ab+/++c
1).
If single crossover occurs between a&b
Normal combination: ab/++
After crossover: a+/+b
a+ progeny= 27
+b progeny = 23+14=37
Total this progeny =64
The recombination frequency between a&b = (number of recombinants/Total progeny) 100
RF = (64/1000)100 = 6.4%
2).
If single crossover occurs between b&+..
Normal combination: b+/+c
After crossover: bc/++
bc tp progeny= 14
++ tp+ progeny = 16
Total this progeny = 30
The recombination frequency between b & + = (number of recombinants/Total progeny) 100
RF = (30/1000)100 = 3%
3).
If single crossover occurs between a&+
Normal combination: a+/+c
After crossover: ac/++
ac tp progeny= 27
++ progeny = 23
Total this progeny = 50
The recombination frequency between a&+ = (number of recombinants/Total progeny) 100
RF = (50/1000)100 = 5%
Recombination frequency (%) = Distance between the genes (cM)
a--------5 cM-----c---------3 cM--------------b
Expected double crossover frequency = (RF between a & c) * (RF between c & b)
= 5% * 3% = 0.0015
As heterozygote order of genes is a+b/+c+, its genotype after double crossover is acb/++. There is no observed double cross over progeny.
The observed double crossover frequency = 0/1000 = 0
Coefficient of Coincidence (COC) = Observed double crossover frequency / Expected double crossover frequency
= 0 / 0.0015
= 0
Interference = 1-COC
=1-0
=1
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