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
Genetics: From Genes to Genomes, 5th edition
- Shown (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_forwardYou 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_forward
- 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, (b) What is the sequence and inter locus distance between these three genes?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_forwardIn 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_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/XwV/v×Xw/Y v/v, Xw/XwV/v×XW/Y V/v.arrow_forwardIn Drosophila, Dichaete (D) is a mutation on chromosome III witha 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. Question: a.) Using the results of this backcross shown in the following table, diagram the cross, showing the genotypes of the parents and offspring of both crosses. b.) What is the sequence and interlocus distance between thesethree genes? Phenotype NumberDichaete 401ebony, pink 389Dichaete, ebony 84pink 96Dichaete, pink 2ebony 3Dichaete, ebony, pink 12wild type 13arrow_forwardMale Drosophila from a true-breeding wild-type stock were irradiated with X-rays and then mated with females from a true-breeding stock carrying the following recessive mutations on the X chromosome: yellow body (y), crossveinless wings (cv), cut wings (ct), singed bristles (sn), and miniature wings (m). These markers are known to map in the order: Recessive alleles: y, cv, ct, sn, m Dominant alleles: y+, cv+, ct+, sn+, m+ y-cv-ct-sn-m у CV ct sn m X-rays х х X ct sn CV у m y+ CV+ ct+ sn+ m+ х X ? Exceptional female: Most of the female progeny of this cross were phenotypically wild type, but one female exhibited ct and sn mutant characteristics. When this exceptional ct sn female was mated with a male from the true-breeding wild-type stock, twice as many females as males appeared among the progeny. a. What is the nature of the X-ray-induced mutation present in the exceptional female? b. Draw the X chromosomes present in the exceptional ct sn female as they would appear during pairing…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_forwardIn Drosophila, a fully heterozygous female with the X-linked recessive genes a, b, and c (not necessarily in that order on the chromosome) was mated to a male that was genetically a, b, c (not necessarily in that order on the chromosome). The offspring occurred in the following phenotypic ratios: Phenotypes: Numbers: What is the cis/trans arrangement in the heterozygous parent? Wild 426 а, с, b 428 Which gene is in the middle? a 23 c, b 22 If you added 23, 22, 3, and 2, it would give you the map distance between genes C 49 b, a 46 What calculation would you make to determine if interference was occurring? (you don't have to complete the calculation) b. C, a Total 1000 3.arrow_forwardIn Drosophila,, the curled mutation (cu, chromosome 3, position 50.0) results in wings that curl up, while ebony (e, chromosome 3, position 70.7) results in a dark body. True breeding, wild type females are mated with true breeding males with curled wings and ebony bodies. Considering Drosophila notation, which of the following correctly diagrams the P1 cross? X X ++ e + + + O+ X + X + ■ + X + + + 3+ X X X X + + Y Y cu cu cu + cu cu J e e e e e (D e + cu cu (Darrow_forward
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