Concept explainers
The recessive, X-linked z1 mutation of the Drosophila gene zeste (z) can produce a yellow (zeste) eye color only in flies that have two or more copies of the wild-type white (w) gene. Using this property, tandem duplications of the w+ gene called w+R were identified. Males with the genotype y+ z1 w+R+spl / Y thus have zeste eyes. These males were crossed to females with the genotype y z1 w+R+ spl / y+ z1 w+R spl+ . (These four genes are closely linked on the X chromosome, in the order given in the genotype, with the centromere to the right of all these genes: y = yellow bodies; y+ = tan bodies; spl = split bristles; spl+ = normal bristles.) Out of 81,540 male progeny of these females, the following exceptions were found:
Class A 2430 yellow bodies, zeste eyes, wild-type bristles
Class B 2394 tan bodies, zeste eyes, split bristles
Class C 23 yellow bodies, wild-type eyes, wild-type bristles
Class D 22 tan bodies, wild-type eyes, split bristles
a. | What were the |
b. | What events gave rise to progeny of classes A and B? |
c. | What events gave rise to progeny of classes C and D? |
d. | On the basis of these experiments, what is the genetic distance between y and spl? |
Trending nowThis is a popular solution!
Chapter 13 Solutions
Genetics: From Genes to Genomes
- Black body (b) and purple eye (pr) are recessive autosomal mutations in Drosophila. Bridges are crossed b/b females with pr/pr males and in the F2 observed 684 wild type, 371 black-bodied, and 300 purple-eyed flies. Do these results indicate that the b and pr genes are closely linked? Explain. (Remember that there is no crossing-over in male Drosophila)..arrow_forwardThe phenotype of vestigial (short) wings (vg) in Drosophila melanogaster is caused by a recessive mutant gene that independently assorts with a recessive gene for hairy (h) body. Assume that a cross is made between a fly that is homozygous for normal wings and has a hairy body and a fly with vestigial wings that is homozygous for normal body. The wild-type F1 flies were crossed among each other to produce 1024 F2 offspring. Which phenotypes would you expect among the F2 offspring, and how many of each phenotype would you expect? Group of answer choices 192 wild type, 256 vestigial, 64 hairy, and 192 vestigial and hairy All vestigial and hairy. 576 wild type, 192 vestigial, 192 hairy, and 64 vestigial and hairy All wild type 256 wild type; 256 vestigial, 256 hairy, and 256 vestigial and hairyarrow_forwardIn Drosophila, Lyra (Ly) and stubble (Sb) are dominant mutations located at locus 40 and 58, on chromosome 3. A recessive mutation with bright eyes was discovered and shown also to be on chromosome 3. A map was obtained by crossing a female who was heterozygous for all three mutations to a male homozygous for the bright red mutation (temporarily will be called br). The following data were obtained. Ly Sb br 404 + + br 2 Ly + br 75 + Sb + 59 Ly + + 18 + Sb br 16 Ly Sb + 4 + + + 422 Diagram the cross and determine the location of the bright red mutation on chromosome 3arrow_forward
- Females of wild-type Strain A and males of mutant Strain B, as well as females of mutant Strain B and males of wild-type Strain A, make reciprocal crosses. Explain why reciprocal crosses are needed in genetics experiments involving Drosophila fruit flies.arrow_forwardIn 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 Offspring sc s v 314 + + + 280 + s v 150 sc + + 156 sc + v 46 + s + 30 sc s + 10 + + v 14 No determination of sex was made in the data. (a) Using proper nomenclature, determine the genotypes of the P1 and F1 parents. (b) Determine the sequence of the three genes and the map distances between them. (c) Are there more or fewer double crossovers than expected? (d) Calculate the coefficient of coincidence. Does it represent positive or negative interference?arrow_forwardIn Drosophila, a cross was made between females expressing thethree X-linked recessive traits, scute bristles (sc), sable body (s),and vermilion eyes (v), and wild-type males. All females were wildtype in the F1, while all males expressed all three mutant traits.The cross was carried to the F2 generation and 1000 offspringwere counted, with the results shown in the following table. Nodetermination of sex was made in the F2 data. Question:Calculate the coefficient of coincidence; does this represent positive or negative interference? Phenotype Offspringsc s v 314+ + + 280+ s v 150sc + + 156sc + v 46+ s + 30sc s + 10+ + v 14arrow_forward
- Male 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_forwardIn Drosophila, a cross was made between females and wild-type males. The female parents expressed the three X-linked recessive traits miniature wing (m), ebony body (e) and cinnabar eye (c). 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 1500 offspring were produced. 412 403 211 Miniature wing, ebony body, cinnabar eye Wild type Ebony body, cinnabar eye Miniature wing Miniature wing, cinnabar eye Ebony body Miniature wing, ebony body Cinnabar eye 224 103 96 23 28 (i) Using proper nomenclature, determine the genotypes of the females and males in generation P and F1. (ii) What is the correct order of the genes? (iii) Construct a genetic map showing the correct order and distances between these genes.arrow_forwardDrosophila females heterozygous for three recessive mutations, a, b, and c , were crossed to males homozygous for all three mutations.The cross yielded the following results: in the image Q. Construct a linkage map showing the correct order of these genes and estimate the distances between them.arrow_forward
- Cinnabar eyes (cn) and reduced bristles (rd) are linked autosomal recesive characters in Drosophila fruit flies. A homozygous wild-type female was crossed to a reduced, cinnabar male, and the F1 females were then crossed with cinnabar reduced males to obtain the F2. Of the 200 F2 offspring obtained, 86 were wild type, 13 were cinnabar, 17 were reduced and 84 were reduced and cinnabar.. What is the map distance between the cn and rd alleles? 8,4 mu 15 mu 17 mu 30 mu 13 muarrow_forwardThree recessive mutations in Drosophila melanogaster, roughoid (ru, small rough eyes), javelin (jv, cylindrical bristles), and sepia eyes (se, dark brown eyes) are linked. A three-point cross was carried out and the following progeny obtained: jv+ ru+ se+ 37 jv+ ru+ se 2 jv+ ru se 14 jv+ ru se+ 146 jv ru se+ 2 jv ru se 35 jv ru+ se 154 jv ru+ se+ 10 a. Determine the order of the genes on the chromosome. b. Determine which progeny contain single crossovers and which contain double crossovers and indicate where among the genes the crossovers occurred. c. Calculate the map distances among the genes. d. Calculate the coefficient of coincidence and interference among the genesarrow_forwardIn fruit flies, you are mapping three genes in a three point cross. The mutants are hairy body (h), sepia colored eyes (se) and female sterility (g). You cross a heterozygous parent with a homozygous recessive parent and obtain the following results: Type Number h se g. 5 + se + 450 + se g 27 ++g_ h se + + + + h + g. h + + TOTAL is the gene in the middle and the distance in map units between se and g is Oh; 16.4 se; 7.1 Oh; 7.1 70 82 7 327 32 1000 se; 16.4arrow_forward
- Human Anatomy & Physiology (11th Edition)BiologyISBN:9780134580999Author:Elaine N. Marieb, Katja N. HoehnPublisher:PEARSONBiology 2eBiologyISBN:9781947172517Author:Matthew Douglas, Jung Choi, Mary Ann ClarkPublisher:OpenStaxAnatomy & PhysiologyBiologyISBN:9781259398629Author:McKinley, Michael P., O'loughlin, Valerie Dean, Bidle, Theresa StouterPublisher:Mcgraw Hill Education,
- Molecular Biology of the Cell (Sixth Edition)BiologyISBN:9780815344322Author:Bruce Alberts, Alexander D. Johnson, Julian Lewis, David Morgan, Martin Raff, Keith Roberts, Peter WalterPublisher:W. W. Norton & CompanyLaboratory Manual For Human Anatomy & PhysiologyBiologyISBN:9781260159363Author:Martin, Terry R., Prentice-craver, CynthiaPublisher:McGraw-Hill Publishing Co.Inquiry Into Life (16th Edition)BiologyISBN:9781260231700Author:Sylvia S. Mader, Michael WindelspechtPublisher:McGraw Hill Education