1.In Drosophila, brittle (b), ebony body (e) and gaucho (g) are three recessive
ID: 53536 • Letter: 1
Question
1.In Drosophila, brittle (b), ebony body (e) and gaucho (g) are three recessive genes. If homozygous EBONY females are crossed with homozygous BRITTLE & GAUCHO males, the resulting F1 progeny are all wild-type. If heterozygous F1 females are mated with EBONY, BRITTLE & GAUCHO males, the following 1,000 progeny appear
a) Derive a map for the three (3) genes.
ebony: 280
brittle, gaucho: 290
brittle, ebony: 8
gaucho: 12
brittle: 120
ebony, gaucho: 110
wild-type: 88
brittle, gaucho, ebony: 92
Explanation / Answer
First we need to identify which type of cross it is. Since there are 8 progeny we know that this is a three point cross.
In order to identify the parental types we need to look the progeny genotypes and look for the genotype which is maximum in number.
In the given case the maximum number genotyes are ebony and brittle, gaucho. Thus, the parental genotypes are ebony and brittle, gaucho
Now we need to identify the double crossovers. The rare or minimum number genotype shows double crossovers.
In the given case, the progeny genotype which shows the lowest number are:
gaucho and brittle, ebony.
By determining gene order we can confirm that we have chosen the right parentals. We compare the two parentals and double crossovers.
ebony and brittle,gaucho (Parentals) .
gaucho and brittle ebony (Double crossovers).
We notice that gene brittle or b remains in the same position i.e. in the middle position in both these crosses. Thus the gene order can be e-b-g or g-b-e. In order to know the gene order for sure as well as map distance we move to the next step.
Calculating the map distances between e and b by using formula we get,
(Total Single crossovers + Double crossovers) X 100/ Total Number of progeny ---->(1)
Substituting values in (1) we get,
(120 + 110 + 8+12) X 100 /1000 = 25000/1000 = 25 map units
Thus distance between e and b is 25 map units.
Considering the recombination frequency between only the region b--g we get,
Now substituting the values in equation (1) we get,
(88+92+8+12) X 100/1000 = 20000/1000 = 20 map units
The map distances are 25 and 20 map units between ebony and brittle, brittle and gaucho respectively.
The total distance between c and b is found to be 25+20= 45 mapunits.
Drawing the map we get,
e---------------b---------------------g
25 20
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