Assume that investigators crossed a strain of fruit flies carrying the dominant
ID: 57932 • Letter: A
Question
Assume that investigators crossed a strain of fruit flies carrying the dominant eye mutation Lobe on the second chromosome with a strain homozygous for the second chromosome recessive mutations smooth abdomen and straw body. The F1 Lobe females were then backcrossed with homozygous smooth abdomen, straw body males, and the following phenotypes were observed: smooth abdomen, straw body 820 Lobe 780 smooth abdomen, Lobe 42 straw body 58 smooth abdomen 148 Lobe, straw body 152
a) Give the gene order and map units between these three loci. Best to show your work. [hint: when a phenotype for a locus in a cross is not listed, assume the individuals are normal or “wild-type” for that trait.]
b) What is the coefficient of coincidence?
Explanation / Answer
Denote smooth body by sA ; straw by sT ; lobe by L
Counter alleles will be SA ; ST ; l
The genotypes and segregation pattern are to be verified:
Parental genotype : those ones with highest frequency
sA l sT (smooth abdomen, straw body) ; SAL ST (Lobe)
Now look which frequencies almost match each other and group them together
We find that smooth abdomen, Lobe and straw body (1)
smooth abdomen and lobe straw (2)
Compare with parents and see which alleles are segregating together.
We get,
So, we have a gene order sA l sT i.e. smooth body – lobe – straw body
Linkage map is calculated from recombination frequency:
Recombination frequency between smooth body and lobe
= no. of recombinants among them / total progeny
= 42 + 58 / 2000
= 0.05 or 5 % or 5mU
Recombination frequency between lobe and straw body
= 148 + 152 / 2000
= 0.15 or 15 % or 15 mU
So, total distance between smooth body and straw body is 20 mU.
b.
Coefficient of coincidence (c.o.c) =
observed frequency of double recombinants / expected frequency of double recombinants
Expected frequency of double recombinants = (.05 x .15) x 2000
= 15
Observed = 0
c.o.c = infinite or you can say zero
Interference = 1 – c.o.c
= 1 – 0
= 1 .0 or 100 %
It means there are no double cross overs.