The Drosophila chromosome 4 is extremely small; virtually no recombination occurs between genes on this chromosome. You have available three differently marked chromosome 4s: one has a recessive allele of the gene eyeless (ey), causing very small eyes; one has a recessive allele of the cubitus interruptus (ci) gene, which causes disruptions in the veins on the wings; and the third carries recessive alleles of both genes. Drosophila adults can survive with two or three, but not with one or four, copies of chromosome 4.
a. | How could you use these three chromosomes to find Drosophila mutants with defective meioses causing an elevated rate of nondisjunction? |
b. | Would your technique allow you to discriminate nondisjunction occurring during the first meiotic division from nondisjunction occurring during the second meiotic division? |
c. | What progeny types would you expect if a fly recognizably formed from a gamete produced by nondisjunction were testcrossed to a fly homozygous for a chromosome 4 carrying both ey and ci? |
d. | Geneticists have isolated so-called compound 4th chromosomes in which two entire chromosome 4s are attached to the same centromere. How can such chromosomes be used to identify mutations causing increased meiotic nondisjunction? Are there any advantages relative to the method you described in part (a)? |
a.
To determine:
The way to use the three chromosomes to identify the mutant strains of Drosophila with defective meiotic cell divisions that resulted in elevated nondisjunction rate.
Introduction:
The three marked fourth chromosomes can be depicted as ci+ ey, ci ey+, and ci ey. Drosophila can survive with two or three copies of chromosome 4, but not with single or four copies.
Explanation of Solution
Mate the potential meiotic mutants having genotype ci+ ey/ci ey+ with homozygotes having genotype ey ci/ey ci. The normal segregants should be ci ey+/ey ci and ci+ ey/ey ci. In meiosis I, nondisjunction will be seen as the progeny having genotype ci+ ey/ci ey+/ey ci. The null-4 gametes that do not have any copy of chromosome number 4 would form zygotes with only a single copy of chromosome 4 that would not survive.
b.
To determine:
Whether the technique discussed in part (a) would allow discriminating nondisjunction occurring during meiosis I from nondisjunction during meiosis II.
Introduction:
The genotype of potential meiotic mutants will be ci+ ey/ci ey+ and homozygotes will be ey ci/ey ci.
Explanation of Solution
The mating between the potential meiotic mutants having genotype ci+ ey/ci ey+ and homozygotes having genotype ey ci/ey ci will detect nondisjunction. However, this method will not differentiate between nondisjunction occurring during meiosis I and nondisjunction during meiosis II.
c.
To determine:
The progeny types formed by the crossing between a fly that developed from a gamete produced by nondisjunction and homozygote fly.
Introduction:
The testcross can be depicted as
Explanation of Solution
In a trisomic fly there are three different ways to pair the chromosome 4. The first option involves
d.
To determine:
The way by which compound 4th chromosomes can be used to identify mutations and its advantages relative to the method described in part (a).
Introduction:
The genotype of a fly with attached fourth chromosomes that are not marked can be depicted as
Explanation of Solution
The compound 4th chromosomes can be used in crosses to assay potential mutants. For example, in cross between
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Chapter 13 Solutions
Genetics: From Genes To Genomes (6th International Edition)
- In Drosophila, vermilion eye color is due to a recessive allele (v) located on the X chromosome. Curved wings are due to a recessive allele (cu) located on one autosome, and ebony body is due to a recessive allele (e) located on another autosome. A vermilion male is mated to a curved, ebony female, and the F1 males are phenotypically wild-type. If these males were backcrossed to curved, ebony females, what proportion of the F2 offspring will be wild-type males?arrow_forwardAnother recessive mutation in Drosophila, ebony (e), is on anautosome (chromosome 3) and causes darkening of the bodycompared with wild-type flies. What phenotypic F1 and F2 maleand female ratios will result if a scalloped-winged female withnormal body color is crossed with a normal-winged ebony male?Work this problem by both the Punnett square method and theforked-line method.arrow_forwardOne of the X chromosomes in a particular Drosophila female had a normal order of genes but carried recessive alleles of the genes for yellow body color (y), vermilion eye color (v), and forked bristles (f), as well as the dominant X-linked Bar eye mutation (B). Her other X chromosome carried the wild-type alleles of all four genes, but the region including y+, v+, and f+ (but not B+) was inverted with respect to the normal order of genes. This female was crossed to a wild- type male in the cross diagrammed her. The cross produced the following male offspring:* table in figure a. Why are there no male offspring with the allele combinations y v f+, v+ v+ f, y v+ f+, or y+ v f (regardless of the allele of the Bar eye gene)?b. What kinds of crossovers produced the y v f b+ and v+ y+ f+ B offspring? Can you determine any genetic distances from these classed of progeny?c. What kinds of crossovers produced the y+ v f+ B+ and y v+ f B offspring?arrow_forward
- One of the X chromosomes in a particular Drosophila female had a normal order of genes but carried recessive alleles of the genes for yellow body color (y), vermilion eye color (v), and forked bristles (f), as well as the dominant X-linked Bar eye mutation (B). Her other X chromosome carried the wild-type alleles of all four genes, but the region including y+, v+, and f+ (but not B+) was inverted with respect to the normal order of genes. This female was crossed to a wild-type male in the cross diagrammed her. The cross produced the following male offspring: Y v f B 48 y+ v+ f+ B+ 45 y v f B+ 11 y+ v+ f+ B 8 y v f B 1 y+ v+ f+ B+ 1 a. Why are there no male offspring with the allele combinations y v f+, v+ v+ f, y v+ f+, or y+ v f (regardless of the allele of the Bar eye gene)? b.What kinds of crossovers produced the y v f b+ and v+ y+ f+ B offspring? Can you determine any genetic distances from these classed of progeny? c. What kinds of crossovers produced the…arrow_forwardDrosophila, yellow body color is due to an X-linked gene that is recessive to the gene for gray body color.a. A homozygous gray female is crossed with a yellow male. The F1 are intercrossed to produce the F2. Give the genotypes and phenotypes, along with the expected proportions, of the F1 and F2 progeny.b. A yellow female is crossed with a gray male. The F1 are intercrossed to produce the F2. Give the genotypes and phenotypes, along with the expected proportions, of the F1 and F2 progeny.c. A yellow female is crossed with a gray male. The F1 females are backcrossed with gray males. Give the genotypes and phenotypes, along with the expected proportions, of the F2 progeny.d. If the F2 flies in part b mate randomly, what are the expected phenotypes and proportions of flies in the F3?arrow_forwardConsider the following variations in Drosophila melanogaster, relative to the wild-type: White eyes are a recessive trait—the gene of which is found in Chromosome I (X). Vestigial wings are a recessive trait—the gene of which is found in Chromosome II. Aristapedia is a dominant trait—the gene of which is found in Chromosome III. Being homozygous for this condition is lethal. Cross the following mutant females with a wild-type (homozygous) male. Show the Punnett square and obtain the genotypic and phenotypic ratios of the first filial generation (F1). Female with white eyes Q4: Show the Punnett squares and obtain the genotypic and phenotypic ratios of the first filial generation (F1) and second filial generation (F2) of the following crosses. Note: The F2 generation can be obtained by crossing one male and one female from the F1 generation. Female with white eyes and vestigial wings and wild-type malearrow_forward
- Female Drosophila heterozygous for three recessive mutations e (ebony body), st (scarlet eyes), and ss (spineless bristles) were testcrossed, and the following progeny were obtained: Phenotype Number wild-type 67 ebony 8 ebony, scarlet 68 ebony, spineless 347 ebony, scarlet, spineless 78 scarlet 368 scarlet, spineless 10 spineless 54 (a) What indicates that the genes are linked? (b) What was the genotype of the original heterozygous females? (c) What is the order of the genes? (d) What is the map distance between e and st? (e) Between e and ss? (f) What is the coefficient of coincidence? (g) Diagram the crosses in this experiment.arrow_forwardIn Drosophila, two mutations, Stubble (Sb) and curled (cu), are linked on chromosome III. Stubble is a dominant gene that is lethal in a homozygous state, and curled is a recessive gene. If a female of the genotype Sb cu + + is to be mated to detect recombinants among her offspring, what male genotype would you choose as a mate?arrow_forwardIt is assumed that in Drosophila the following genotypes produce phenotypes. َA- B- = Red color A- bb = Plum color aa B- = Magenta color aa bb = White color The third latent genotype, cc, kills homozygous Plums, but has no effect on other genotypes. Also, genotype C- does not produce a large phenotype. If first-generation Drosophilas are heterozygous for all of these genes and interbreed, what phenotypic ratios are expected in society?arrow_forward
- The genes dumpy (dp), clot (cl), and apterous (ap) are linked on chromosome II of Drosophila. In a series of two-point mapping crosses, the following genetic distances were determined. What is the sequence of the three genes? dp–ap 42 dp–cl 3 ap–cl 39arrow_forwardThe genes dumpy wings (dp), clot eyes (cl), and apterous wings (ap) are linked on chromosome II of Drosophila. In a series of two-point mapping crosses, the genetic distances shown below were determined. What is the sequence of the three genes? dp–ap 42 dp–cl 3 ap–cl 39arrow_forwardTwo different strains of Drosophila, strain A and strain B, each has a recessive mutation that results in abnormally bright red eye color. (Wild type flies have brownish red eye color). When a homozygous strain A fly is crossed with a homozygous B fly, all of the progeny have the dominant wild type eye color. The wild type-eyed progeny were allowed to breed among themselves to produce the F2 generation. The F2 generation consisted of 92 wild type and 74 bright red-eyed flies. Write the genotype(s) of the flies in each generation. Use a low dash (e.g. A_ B_) to indicate genotypes that could be either homozygous or heterozygous) a) parental strain A b) parental strain B c) wild type progeny (F1) d) wild type F2 e) bright-eyed F2arrow_forward
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