The Drosophila chromosome 4 is extremely small; virtually no recombination occurs between genes on this chromosome. You have available three differently marked chromosome 4’s: 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 12 Solutions
EBK GENETICS: FROM GENES TO GENOMES
- In Drosophila, the genes for withered wings (whd), smooth abdomen (sm) and speck body (sp) are located on chromosome 2 and are separated by the following map distances:whd ----------(30.5)----------sm-----(15.5)-----spA female with withered wings and a smooth abdomen was mated to a male with a speck body.The resulting phenotypically wild-type females were mated with males that had the mutant phenotype for all three traits, producing 1000 offspring. Calculate the number of expected double crossover progeny. Express your answer to the nearest whole number.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_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
- 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:* 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_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: 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_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
- Drosophila, 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_forwardFemale 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_forwardThe 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_forward
- In the fruit fly Drosophila melanogaster, a recessive condition called eyeless (ey) significantly interferes with normal eye development. Eyes are either very small or absent. The ey gene is found on chromosome 4, which is the smallest of the four chromosomes in the organism. Individuals with a single copy or three copies of chromosome 4 are viable and can reproduce. A trisomic wild-type male is crossed with a disomic wild-type female: see attached image Determine the chromosome constitution and phenotypic ratios in the offspring. Assume random segregation of chromosomes into gametes.arrow_forwardThe genes for the recessive traits of mahogany eyes and ebony body are approximately 30 map units apart on chromosome III in Drosophila. Assume that a mahogany-eyed female was mated to an ebony-bodied male and that the resulting F1 females (all phenotypically wild-type) were then mated to mahogany, ebony males. Of 1000 offspring, what would be the expected phenotypes, and in what numbers would they be expected? Group of answer choices None of these is correct wild-type = 200; mahogany and ebony = 200; mahogany = 300; ebony = 300 mahogany = 200; ebony = 200; wild-type = 300; mahogany and ebony = 300 wild-type = 350; mahogany and ebony = 350; mahogany = 150; ebony = 150 mahogany = 350; ebony = 350; wild-type = 150; mahogany and ebony = 150arrow_forwardVermillion eye color in Drosophila sp. is a sex-linked recessive trait. What phenotype would be found in this progeny of a cross between a vermillion female and a wild type male?arrow_forward
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