Concept explainers
In Drosophila, the autosomal recessive dp allele of the dumpy gene produces short, curved wings, while the autosomal recessive allele bw of the brown gene causes brown eyes. In a testcross using females heterozygous for both of these genes, the following results were obtained: wild-type wings, wild-type eyes 178 wild-type wings, brown eyes 185 dumpy wings, wild-type eyes 172 dumpy wings, brown eyes 181
In a testcross using males heterozygous for both of these genes, a different set of results was obtained: wild-type wings, wild-type eyes 247 dumpy wings, brown eyes 242
a. | What can you conclude from the first testcross? |
b. | What can you conclude from the second testcross? |
c. | How can you reconcile the data shown in parts (a) and (b)? Can you exploit the difference between these two sets of data to devise a general test for synteny in Drosophila? |
d. | The genetic distance between dumpy and brown is 91.5 m.u. How could this value be measured? |
Want to see the full answer?
Check out a sample textbook solutionChapter 5 Solutions
Genetics: From Genes to Genomes
Additional Science Textbook Solutions
Becker's World of the Cell (9th Edition)
Evolutionary Analysis (5th Edition)
Microbiology with Diseases by Body System (4th Edition)
Loose Leaf For Integrated Principles Of Zoology
Marine Biology (Botany, Zoology, Ecology and Evolution)
Biology: Concepts and Investigations
- Hemophilia and color blindness are both recessive conditions caused by genes on the X chromosome. To calculate the recombination frequency between the two genes, you draw a large number of pedigrees that include grandfathers with both hemophilia and color blindness, their daughters (who presumably have one chromosome with two normal alleles and one chromosome with two mutant alleles), and the daughters sons. Analyzing all the pedigrees together shows that 25 grandsons have both color blindness and hemophilia, 24 have neither of the traits, 1 has color blindness only, and 1 has hemophilia only. How many centimorgans (map units) separate the hemophilia locus from the locus for color blindness?arrow_forwardAs it turned out, one of the tallest Potsdam Guards had an unquenchable attraction to short women. During his tenure as guard, he had numerous clandestine affairs. In each case, children resulted. Subsequently, some of the childrenwho had no way of knowing that they were relatedmarried and had children of their own. Assume that two pairs of genes determine height. The genotype of the 7-foot-tall Potsdam Guard was A9A9B9B9, and the genotype of all of his 5-foot clandestine lovers was AABB. An A9 or B9 allele in the offspring each adds 6 inches to the base height of 5 feet conferred by the AABB genotype. a. What were the genotypes and phenotypes of all the F1 children? b. Diagram the cross between the F1 offspring, and give all possible genotypes and phenotypes of the F2 progenyarrow_forwardIn Drosophila, gray body color is dominant to ebony body color, while long wings are dominant to vestigial wings. Assuming that the P1 individuals are homozygous, work the following crosses through the F2 generation, and determine the genotypic and phenotypic ratios for each generation. (a) gray, long * ebony, vestigial (b) gray, vestigial * ebony, long (c) gray, long * gray, vestigialarrow_forward
- Female fruit flies from a true-breeding strain with yellow body and cut wings are mated to male fruit flies from a true breeding strain that is wild type for both body color and wings. In the F1 generation, all of the female flies are wild type for both characters while all of the male flies have yellow bodies and cut wings. The F1 males and females are intercrossed, and among 1000 male progeny the following is observed: 425 yellow body and cut wings 405 wild type body and wild type wings 90 yellow body and wild type wings 80 wild type body and cut wings Are these genes X-linked or autosomal? How far apart are they from each other?arrow_forwardA cross was performed using Drosophila melanogaster involving a female known to be heterozygous for both ebony body and sepia eyes and a male known to be homozygous for both of these recessive traits. The following data was produced from the cross. Test these data to determine if they are significantly different from the expected phenotypic ratio. Remember to use the 5% level of significance Wild eye Wild body – 102, Wild eye Ebony body – 94, Sepia eye Wild body – 100, Sepia eye Ebony body – 93. Your answer should include the hypothesized cross in genotypes, the Chi-squared value, the critical value and whether you reject or do not reject.arrow_forwardIn Drosophila, males from a true-breeding stock with raspberry-colored eyes were mated to females from a true-breeding stock with sable-colored bodies. In the F1 generation, all the females had wild-type eye and body color, while all the males had wild-type eye color but sable-colored bodies. When F1 males and females were mated to each other, the F2 was composed of: 216 females with wild-type eyes and wild-type bodies 223 females with wild-type eyes and sable bodies 191 males with wild-type eyes and sable bodies 188 males with raspberry eyes and wild-type bodies 23 males with wild-type eyes and bodies 27 males with raspberry eyes and sable bodies Which statements are consistent with the above data? (Select all correct answers.) The alleles causing the raspberry-colored eye and sable-colored body phenotypes are dominant to the corresponding wild-type alleles The genes controlling raspberry-colored eyes and sable-colored bodies map…arrow_forward
- . In fruit flies (Drosophila melanogaster), the bn+ allele for normal dull red eyes is dominant to the bn allele that gives brown eyes. Another gene affects wing shape; for this gene, the ct+ allele for normal wings is dominant to the ct allele, which gives “cut” wings, with jagged edges. A fly with dull red eyes and normal wings was crossed with a fly that had dull red eyes and cut wings, and the following progeny were obtained: 16 dull red eyes, normal wings 14 dull red eyes, cut wings 5 brown eyes, normal wings 5 brown eyes, cut wings What were the genotypes of the parents?arrow_forward. In mice, the following alleles were used in a cross: W = waltzing gait w = nonwaltzing gait G = normal gray color g = albino B = bent tail b = straight tail A waltzing gray bent-tailed mouse is crossed with a nonwaltzing albino straight-tailed mouse and, over several years, the following progeny totals are obtained: waltzing gray bent 18 waltzing albino bent 21 nonwaltzing gray straight 19 nonwaltzing albino straight 22 waltzing gray straight 4 waltzing albino straight 5 nonwaltzing gray bent 5 nonwaltzing albino bent 6 Total 100 a. What were the genotypes of the two parental mice in the cross? b. Draw the chromosomes of the parents.c. If you deduced linkage, state the map unit value or values and show how they were obtained.arrow_forwardIn Drosophila, gray body color is dominant to ebony body color, while long wings are dominant to vestigial wings. Assuming that the P1 individuals are homozygous, work the following crosses through the F2 generation, and determine the genotypic and phenotypic ratios for each generation. (a) gray, long x ebony, vestigialarrow_forward
- In mice, the trait for high cholesterol is specified by a dominant allele designated HC, whereas the wild-type allele for normal cholesterol levels is designated hc. Black fur is specified by a recessive allele designated bl, whereas the wild-type allele which gives brown fur is designated BL. The genes for both of these traits are 30cM apart on the same autosome. A brown female (#1) with high cholesterol is mated to a black male (#2) with normal cholesterol. The progeny from this cross include a brown male (#3) with high cholesterol and a black female (#4) with normal cholesterol. What is the probability that the black mouse in the progeny of the first cross will also have high cholesterol?arrow_forwardIn fruit flies, long wings (M) is dominant to miniature wings (m) and red eyes (B) is dominant to brown eyes (b). You testcross a long-winged, red-eyed fly (genotype Mm Bb) female to a mini-winged, brown-eyed male (genotype mm bb) and get the following progeny: 256 long wings, red eyes 261 mini wings, red eyes 240 long wings, brown eyes 243 mini wings, brown eyes Are these genes for wing length and eye color linked or unlinked? How do you know?arrow_forwardIn the fruit fly Drosophila melanogaster, the following genes and mutations are known:Wing size: recessive allele for tiny wings t; dominantallele for normal wings T.Eye shape: recessive allele for narrow eyes n;dominant allele for normal (oval) eyes N.For each of the four following crosses, give thegenotypes of each of the parents.Male FemaleWings Eyes Wings Eyes Offspring1 tiny oval × tiny oval 78 tiny wings, oval eyes24 tiny wings, narrow eyes2 normal narrow × tiny oval 45 normal wings, oval eyes40 normal wings, narrow eyes38 tiny wings, oval eyes44 tiny wings, narrow eyes3 normal narrow × normal oval 35 normal wings, oval eyes29 normal wings, narrow eyes10 tiny wings, oval eyes11 tiny wings, narrow eyes4 normal narrow × normal oval 62 normal wings, oval eyes19 tiny wings, oval eyesarrow_forward
- Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning