4. Genetic Mapping and Linkage

Mapping Genes

Open Question
In the cross shown here, involving two linked genes, ebony (e) and claret (ca), in Drosophila, where crossing over does not occur in males, offspring were produced in a 2 + : 1 ca : 1 e phenotypic ratio:These genes are 30 units apart on chromosome III. What did crossing over in the female contribute to these phenotypes?
Open Question
In a series of two-point mapping crosses involving five genes located on chromosome II in Drosophila, the following recombinant (single-crossover) frequencies were observed:pr–adp 29%pr–vg 13pr–c 21pr–b 6adp–b 35adp–c 8adp–vg. 16vg–b. 19vg–c 8c–b. 27In another set of experiments, a sixth gene, d, was tested against b and pr:d–b 17%d–pr 23%Predict the results of two-point mapping between d and c, d and vg, and d and adp.
Open Question
In a series of two-point mapping crosses involving five genes located on chromosome II in Drosophila, the following recombinant (single-crossover) frequencies were observed:pr–adp 29%pr–vg 13pr–c 21pr–b 6adp–b 35adp–c 8adp–vg. 16vg–b. 19vg–c 8c–b. 27Given that the adp gene is near the end of chromosome II (locus 83), construct a map of these genes.
Open Question
In Drosophila, a cross was made between females—all expressing the three X-linked recessive traits scute bristles (sc), sable body (s), and vermilion eyes (v)—and wild-type males. In the F₁, all females were wild type, while all males expressed all three mutant traits. The cross was carried to the F₂ generation, and 1000 offspring were counted, with the results shown in the following table.Phenotype Offspringsc s v 314+ + + 280+ s v 150sc + + 156sc + v 46+ s + 30sc s + 10+ + v 14No determination of sex was made in the data.Determine the sequence of the three genes and the map distances between them.
Open Question
In Drosophila, a cross was made between females—all expressing the three X-linked recessive traits scute bristles (sc), sable body (s), and vermilion eyes (v)—and wild-type males. In the F₁, all females were wild type, while all males expressed all three mutant traits. The cross was carried to the F₂ generation, and 1000 offspring were counted, with the results shown in the following table.Phenotype Offspringsc s v 314+ + + 280+ s v 150sc + + 156sc + v 46+ s + 30sc s + 10+ + v 14No determination of sex was made in the data.Using proper nomenclature, determine the genotypes of the P₁ and F₁ parents.
Open Question
Another cross in Drosophila involved the recessive, X-linked genes yellow (y), white (w), and cut (ct). A yellow-bodied, white-eyed female with normal wings was crossed to a male whose eyes and body were normal but whose wings were cut. The F₁ females were wild type for all three traits, while the F₁ males expressed the yellow-body and white-eye traits. The cross was carried to an F₂ progeny, and only male offspring were tallied. On the basis of the data shown here, a genetic map was constructed.Phenotype Male Offspringy + ct 9+ w + 6y w ct 90+ + + 95+ + ct 424y w + 376y + + 0+ w ct 0Could the F₂ female offspring be used to construct the map? Why or why not?
Open Question
Another cross in Drosophila involved the recessive, X-linked genes yellow (y), white (w), and cut (ct). A yellow-bodied, white-eyed female with normal wings was crossed to a male whose eyes and body were normal but whose wings were cut. The F₁ females were wild type for all three traits, while the F₁ males expressed the yellow-body and white-eye traits. The cross was carried to an F₂ progeny, and only male offspring were tallied. On the basis of the data shown here, a genetic map was constructed.Phenotype Male Offspringy + ct 9+ w + 6y w ct 90+ + + 95+ + ct 424y w + 376y + + 0+ w ct 0Construct a map, assuming that white is at locus 1.5 on the X chromosome.
Open Question
Another cross in Drosophila involved the recessive, X-linked genes yellow (y), white (w), and cut (ct). A yellow-bodied, white-eyed female with normal wings was crossed to a male whose eyes and body were normal but whose wings were cut. The F₁ females were wild type for all three traits, while the F₁ males expressed the yellow-body and white-eye traits. The cross was carried to an F₂ progeny, and only male offspring were tallied. On the basis of the data shown here, a genetic map was constructed.Phenotype Male Offspringy + ct 9+ w + 6y w ct 90+ + + 95+ + ct 424y w + 376y + + 0+ w ct 0Diagram the genotypes of the F₁ parents.
Open Question
In Drosophila, Dichaete (D) is a mutation on chromosome III with a dominant effect on wing shape. It is lethal when homozygous. The genes ebony body (e) and pink eye (p) are recessive mutations on chromosome III. Flies from a Dichaete stock were crossed to homozygous ebony, pink flies, and the F₁ progeny, with a Dichaete phenotype, were backcrossed to the ebony, pink homozygotes. Using the results of this backcross shown in the table,Phenotype NumberDichaete 401ebony, pink 389Dichaete, ebony 84pink 96Dichaete, pink 2ebony 3Dichaete, ebony, pink 12wild type 13What is the sequence and interlocus distance between these three genes?
Open Question
The table given here lists the arrangement of alleles of linked genes in dihybrid organisms, the recombination frequency between the genes, and specific gamete genotypes. Using the information provided, determine the expected frequency of the listed gametes. Assume one map unit equals 1% recombination and, when three genes are involved, interference is zero.

Dihybrid Recombination Gamete
Genotype Frequency Genotype
A. DE/de 8% De
B. AD/ad 28% ad
C. DEF/def E–F 24% dEf
D–E 8%
D. BdE/bDe B–D 18% Bde
D–E 8%
Open Question
Drosophila females homozygous for the third chromosomal genes pink and ebony (the same genes from Problem 16) were crossed with males homozygous for the second chromosomal gene dumpy. Because these genes are recessive, all offspring were wild type (normal). F₁ females were testcrossed to triply recessive males. If we assume that the two linked genes, pink and ebony, are 20 mu apart, predict the results of this cross. If the reciprocal cross were made (F₁ males—where no crossing over occurs—with triply recessive females), how would the results vary, if at all?
Open Question
The Rh blood group in humans is determined by a gene on chromosome 1. A dominant allele produces Rh+ blood type, and a recessive allele generates Rh-. Elliptocytosis is an autosomal dominant disorder that produces abnormally shaped red blood cells that have a short life span resulting in hereditary anemia. A large family with elliptocytosis is tested for genetic linkage of Rh blood group and the disease. The lod score data below are obtained for the family.

Over what range of θ do lod scores indicate significant evidence in favor of genetic linkage? <>
Open Question
The Rh blood group in humans is determined by a gene on chromosome 1. A dominant allele produces Rh+ blood type, and a recessive allele generates Rh-. Elliptocytosis is an autosomal dominant disorder that produces abnormally shaped red blood cells that have a short life span resulting in hereditary anemia. A large family with elliptocytosis is tested for genetic linkage of Rh blood group and the disease. The lod score data below are obtained for the family.

What is Zₘₐₓ for this family?
Open Question
The Rh blood group in humans is determined by a gene on chromosome 1. A dominant allele produces Rh+ blood type, and a recessive allele generates Rh-. Elliptocytosis is an autosomal dominant disorder that produces abnormally shaped red blood cells that have a short life span resulting in hereditary anemia. A large family with elliptocytosis is tested for genetic linkage of Rh blood group and the disease. The lod score data below are obtained for the family.

From these data, can you conclude that Rh and elliptocytosis loci are genetically linked in this family? Why or why not?
Open Question
Genetic linkage mapping for a large number of families identifies 4% recombination between the genes for Rh blood type and elliptocytosis (see Problem 18). At the Rh locus, alleles R and r control Rh+ and Rh- blood types. Allele E producing elliptocytosis is dominant to the wild-type recessive allele e. Tom and Terri each have elliptocytosis, and each is . Tom's mother has elliptocytosis and is Rh- while his father is healthy and has Rh+. Terri's father is Rh+ and has elliptocytosis; Terri's mother is Rh- and is healthy.

What is the probability that a child of Tom and Terri who is Rh+ will have elliptocytosis?