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
- In 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_forwardShown (the picture) is a partial gene map for Drosophila melanogaster. In Drosophila, vermillion eyes are recessive (v) to red eyes (V) and the gene resides on chromosome 1; dumpy wings (d) are recessive to normal wings (D); and speckled body (s) is recessive to normal body (S). The dumpy wing gene and the speckle body gene are linked on chromosome 2 and have 7 map units separating them. Assume that chromosome 1 and 2 are autosomal chromosomes and that there are no known interactions between the genes. In an experiment, an undergraduate student crossed pure breeding males that have red eyes, normal wings, and normal body with pure breeding females who have vermillion eyes, dumpy wings, and speckle body to obtain an F1. Using the Drosophila gene map, answer the following questions: a. What fraction of the F1 gametes will contain the alleles for red eyes, normal wings and speckle body? b. In a cross between an F1 flies and a tester, what fraction of the offspring do you expect to have…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 as diagrammed here. 1. Draw the meiotic prophase I chromosomes with maximal pairing from the inversion heterozygous female. 2. (continued) When a crossover occurs between v and f, please draw the resulting chromosomes in four gametes.arrow_forward
- You are studying a group of mutations in Drosophila all of which are recessive lethal and map to a small area of chromosome 3. You decide to perform complementation testing to see which mutations are in the same gene. You have bred your flies such that they have the following genotype: +; +; TM3 Sb/mut; + This indicates that chromosomes X, 2, and 4 are wild-type. One copy of chromosome 3 is the balancer TM3 Sb which carries a recessive lethal gene and the dominant marker Stubble (Sb), which causes the flies to have short bristles. The other copy of chromosome 3 carries your recessive lethal mutation. a) Describe how you would maintain each of your individual stocks. What ratios of phenotypes do you expect? b) When you perform your complementation testing, you will have to look at the F1 phenotypes to determine whether two genes complement. Describe what your F1 fly population should look like if the two mutations complement and if they do not. You breed each of your four…arrow_forwardIn 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, (a) Diagram this cross, showing the genotypes of the parents and offspring of both crosses. (b) What is the sequence and inter locus distance between these three genes?arrow_forwardIn Drosophila melanogaster, red eyes are dominant over white and the variation for this characteristic is on the X chromosome. Vestigial wings (v) are recessive to normal (V) for an autosomal gene. Predict the appearance of offspring of the following crosses: XW/XwV/v×Xw/Y v/v, Xw/XwV/v×XW/Y V/v.arrow_forward
- In Drosophila melanogaster, red eyes are dominant over white and the variation for this characteristic is on the X chromosome. Vestigial wings (v) are recessive to normal (V) for an autosomal gene. Predict the appearance of offspring of the following crosses: XW/Xw V/v×Xw/Y v/v, Xw/Xw V/v × XW/Y V/v.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_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_forward
- The genotype of a Drosophila with a heterozygous translocation between chromosome 2 and chromosome 3 is shown below, where bw = brown eyes and e = ebony body: bw+ bw e Assume there is no crossing over in the female and that alternate = adjacent 1 = adjacent 2 disjunction. [Yes, these assumptions are incorrect, but they make the problem much easier!] Two individuals of the above genotype, i.e. both heterozygous for this 2;3 translocation and both heterozygous for bw (on chromosome 2) and e (on chromosome 3) are crossed. A (Only zygotes with balanced genomes will survive to adulthood. What proportion of F1 zygotes have a balanced genome? B. Draw the genotype(s) and state the phenotype(s) of the surviving progeny you mentioned in part A above. For each different genotype/phenotype indicate it's frequency among the surviving progeny.arrow_forwardImagine Drosophila genes C, D, and E are autosomal genes located close to each other on the same chromosome (same assumptions as the first problem). You cross a C D E homozygote with a c d e homozygote, then cross the F1 females with a c d e homozygous male. Of 400 progeny, you observe the following phenotypes: 135 CDE 139 cde 22 cDE 18 Cde 42 CdE 38 cDe 3 cdE 3 CDe a. What is the order of the genes? Calculate the distance between genes and draw a map to get the final answer. Question 2 options: 1.CED 2. DEC 3. ECD 4. DCEarrow_forwardThe following corn loci are on one arm of chromosome9 in the order indicated (the distances between themare shown in map units):c-bz-wx-sh-d-centromere12 8 10 20 10C gives colored aleurone; c, white aleurone.Bz gives green leaves; bz, bronze leaves.Wx gives starchy seeds; wx, waxy seeds.Sh gives smooth seeds; sh, shrunken seeds.D gives tall plants; d, dwarf.A plant from a standard stock that is homozygous for allfive recessive alleles is crossed with a wild-type plantfrom Mexico that is homozygous for all five dominantalleles. The F1 plants express all the dominant allelesand, when backcrossed to the recessive parent, give thefollowing progeny phenotypes:colored, green, starchy, smooth, tall 360white, bronze, waxy, shrunk, dwarf 355colored, bronze, waxy, shrunk, dwarf 40white, green, starchy, smooth, tall 46colored, green, starchy, smooth, dwarf 85white, bronze, waxy, shrunk, tall 84colored, bronze, waxy, shrunk, tall 8white, green, starchy, smooth, dwarf 9colored, green, waxy, smooth,…arrow_forward
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