11. Fruit-fly geneticists generated a trihybrid by performing a cross between an
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11. Fruit-fly geneticists generated a trihybrid by performing a cross between an inbred wild-type line and a line that is homozygous for three recessive mutant alleles: black body (b), dumpy wings (dp) and purple eyes (pr) They then crossed Fi trihybrid females with males from the triple homozygous recessive line. By observing the phenotypes of 2000 progeny from these F1 crosses, they inferred the inheritance of the following haploid genotypes from the trihybrid. Gamete Genotype Count b* dp pr 593 b dp pr 51 b' dp pr 335 b pr dp b dp pr 18 b dp pr 35 cM 7 CM b dp pr 325 b dp pr 607 b dp pr 2000 Total In the space above, draw a map showing the order of the three genes on the chromosome. Label the map to indicate the distances between the gene in the middle and the genes on either side of it. Be clear in your labeling whether your distances are expressed as recombination frequencies or centimorgans [10Explanation / Answer
To solve the problem let us go in a stepwise fashion:
.First let us determine which of the genotypes are parental genotypes. The genotypes found most frequently i.e. in maximum numbers are the parental genotypes.
So, from the above table clearly b+dp+pr+ = 593
b dp pr =607 are the parental genotypes.
Next we determine the order of the genes. Once we have determined the parental genotypes, we use that information along with the information obtained from the double cross-over. The double crossover genotypes are always in the lowest number.
So, b+ dp pr =18
b dp+ pr+=22 are the double crossover genotypes.
Next important point is that double crossover event moves the middle allele from one sister chromatid to the other.This effectively places the non-parental allele of the middle gene onto a chromosome with the parental alleles of the two flanking genes. From the information in the question, b gene must be in the middle because the b+ is now on the same chromosome as dp &pr.
So, the order of the genes will be dp b pr.
Now let us determine the distance between the genes.
b+ dp pr+ =335
b dp+ pr =325 These two genotypes represent single crossover between b and dp.
b+ dp+ pr = 51
b dp pr+ =49 These two genotypes represent single crossover between b and pr
So map distance in centi morgan between b and dp will be,
= single crossover genotypes between b and dp + double crossover × 100 cM
Total progeny
= 335+325+18+22 ×10 cM = 35 cM
2000
Similarly, distance between b and pr
= single crossover genotypes between b and dp + double crossover × 100 cM
Total progeny
= 51 +49+18+22 ×100 = 7cM
2000
SO, map will be like
dp 35cM b 7cM pr
To solve the problem let us go in a stepwise fashion:
.First let us determine which of the genotypes are parental genotypes. The genotypes found most frequently i.e. in maximum numbers are the parental genotypes.
So, from the above table clearly b+dp+pr+ = 593
b dp pr =607 are the parental genotypes.
Next we determine the order of the genes. Once we have determined the parental genotypes, we use that information along with the information obtained from the double cross-over. The double crossover genotypes are always in the lowest number.
So, b+ dp pr =18
b dp+ pr+=22 are the double crossover genotypes.
Next important point is that double crossover event moves the middle allele from one sister chromatid to the other.This effectively places the non-parental allele of the middle gene onto a chromosome with the parental alleles of the two flanking genes. From the information in the question, b gene must be in the middle because the b+ is now on the same chromosome as dp &pr.
So, the order of the genes will be dp b pr.
Now let us determine the distance between the genes.
b+ dp pr+ =335
b dp+ pr =325 These two genotypes represent single crossover between b and dp.
b+ dp+ pr = 51
b dp pr+ =49 These two genotypes represent single crossover between b and pr
So map distance in centi morgan between b and dp will be,
= single crossover genotypes between b and dp + double crossover × 100 cM
Total progeny
= 335+325+18+22 ×10 cM = 35 cM
2000
Similarly, distance between b and pr
= single crossover genotypes between b and dp + double crossover × 100 cM
Total progeny
= 51 +49+18+22 ×100 = 7cM
2000
SO, map will be like
dp 35cM b 7cM pr
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