Genetics: From Genes to Genomes
6th Edition
ISBN: 9781259700903
Author: Leland Hartwell Dr., Michael L. Goldberg Professor Dr., Janice Fischer, Leroy Hood Dr.
Publisher: McGraw-Hill Education
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Textbook Question
Chapter 5, Problem 23P
In Drosophila, the recessive allele mb of one gene causes missing bristles, the recessive allele e of a second gene causes ebony body color, and the recessive allele k of a third gene causes kidney-shaped eyes. (Dominant wild-type alleles of all three genes are indicated with a + superscript.) The three different P generation crosses in the table that follows were conducted, and then the resultant F females from each cross were testcrossed to males that were homozygous for the recessive alleles of both genes in question. The
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
In roses, purple flower color is determined by the dominant P allele, while pp
homozygotes are white. The presence of long stems is determined by the
dominant S allele, while ss homozygotes have short stems. Both mutations are
completely penetrant. A test cross was performed between a rose plant of
unknown genotype with a white flowered, short stemmed rose plant (pp ss) and
the following 200 progeny plants were obtained:
84 white flowers, long stems
16 purple flowers, long stems
82 purple flowers, short stems
18 white flowers, short stems
Select the statements below that are TRUE.
Select 2 correct answer(s)
The P and S genes independently assort during meiosis.
The map distance between P and S is 17 CM.
The genotype of the progeny plants with purple flowers and short stems is
PP ss.
The map distance between P and S is 83 CM.
The homologs in the plant with unknown genotype are p S and Ps.
The homologs in the plant with unknown genotype are PS and p s.
In Drosophila, the sepia mutation (se, chromosome 3, position 26) results in dark brown eyes, while cinnabar (cn, chromosome 2, position 57.5) results in bright orange-red eyes. True breeding, wild type females are mated with true breeding males homozygous recessive for both traits.
Using Drosophila notation, diagram the P1 and F1 crosses.
P1 F1
Fill in the chart with phenotypic ratios that would be expected in the F2 generation. Use the space provided to show your work.
Phenotype
Females
Males
Overall (♀and ♂)
=1 =1
In roses, purple flower color is determined by the dominant P allele, while pphomozygotes are white. The presence of long stems is determined by the dominant S allele, while ss homozygotes have short stems. Both mutations are completely penetrant. A test cross was performed between a rose plant of unknown genotype with a white flowered, short stemmed rose plant (pp ss) and the following 200 progeny plants were obtained:
84 white flowers, long stems
16 purple flowers, long stems
82 purple flowers, short stems
18 white flowers, short stems
Select two statements below that are TRUE.
options:
The P and S genes independently assort during meiosis.
The map distance between P and S is 17 cM.
The genotype of the progeny plants with purple flowers and short stems is PP ss.
The map distance between P and S is 83 cM.
The homologs in the plant with…
Chapter 5 Solutions
Genetics: From Genes to Genomes
Ch. 5 - Choose the phrase from the right column that best...Ch. 5 - a. A Drosophila male from a true-breeding stock...Ch. 5 - Prob. 3PCh. 5 - The Punnett square in Fig. 5.4 shows how Mendels...Ch. 5 - In mice, the dominant allele Gs of the X-linked...Ch. 5 - In Drosophila, males from a true-breeding stock...Ch. 5 - If the a and b loci are 20 m.u. apart in humans...Ch. 5 - CCDD and ccdd individuals were crossed to each...Ch. 5 - In mice, the autosomal locus coding for the...Ch. 5 - In a particular human family, John and his mother...
Ch. 5 - Albino rabbits lacking pigment are homozygous for...Ch. 5 - In corn, the allele A allows the deposition of...Ch. 5 - If the a and b loci are 40 cM apart and an AA BB...Ch. 5 - Write the number of different kinds of phenotypes,...Ch. 5 - A DNA variant has been found linked to a rare...Ch. 5 - Figure 5.7a shows chromosomes during prophase of...Ch. 5 - Figure 5.7b shows bivalents in mouse primary...Ch. 5 - Cinnabar eyes cn and reduced bristles rd are...Ch. 5 - In Drosophila, the autosomal recessive dp allele...Ch. 5 - From a series of two-point crosses, the following...Ch. 5 - Map distances were determined for four different...Ch. 5 - In the tubular flowers of foxgloves, wild-type...Ch. 5 - In Drosophila, the recessive allele mb of one gene...Ch. 5 - A snapdragon with pink petals, black anthers, and...Ch. 5 - In Drosophila, three autosomal genes have the...Ch. 5 - Drosophila females heterozygous for each of three...Ch. 5 - Male Drosophila expressing the autosomal recessive...Ch. 5 - a. In Drosophila, crosses between F1 heterozygotes...Ch. 5 - A true-breeding strain of Virginia tobacco has...Ch. 5 - Prob. 30PCh. 5 - The following list of four Drosophila mutations...Ch. 5 - Do the data that Mendel obtained fit his...Ch. 5 - Two genes control color in corn snakes as follows:...Ch. 5 - A mouse from a true-breeding population with...Ch. 5 - Neurospora of genotype a c are crossed with...Ch. 5 - A cross was performed between one haploid strain...Ch. 5 - Prob. 40PCh. 5 - Prob. 41PCh. 5 - Indicate the percentage of tetrads that would have...Ch. 5 - Prob. 43PCh. 5 - This problem leads you through the derivation of a...Ch. 5 - a. In ordered tetrad analysis, what is the maximum...Ch. 5 - Prob. 46PCh. 5 - A single yeast cell placed on a solid agar will...Ch. 5 - Figure 5.29 shows mitotic recombination leading to...Ch. 5 - A diploid strain of yeast has a wild-type...Ch. 5 - In Drosophila, the yellow y gene is near the...Ch. 5 - Neurofibromas are tumors of the skin that can...Ch. 5 - Two important methods for understanding the...
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biology and related others by exploring similar questions and additional content below.Similar questions
- 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_forwardFigure 8.10 In pea plants, purple flowers (P) are dominant to white (p), and yellow peas (Y) are dominant to green (y). What are the possible genotypes and phenotypes for a cross between PpYY and ppYy pea plants? How many squares would you need to complete a Punnett square analysis of this cross?arrow_forwardIndividuals of genotype AaBb were mated to individuals of genotype aabb. One thousand offspring were counted, with the following results: 474 Aabb, 480 aaBb, 20 AaBb, and 26 aabb. What type of cross is it? Are these loci linked? What are the two parental classes and the two recombinant classes of offspring? What is the percentage of recombination between these two loci? How many map units apart are they?arrow_forward
- In Drosophila, singed bristles (sn) and cut wings (ct) are both caused by recessive, X-linked alleles. The wild type alleles (sn+ and ct+) are responsible for straight bristles and intact wings, respectively. A female homozygous for sn and ct+ is crossed to a sn+ct male. The F1 flies are interbred. The F2 males are distributed as follows: genotype number sn ct 15 sn ct+ 34 sn+ ct 33 sn+ct+ 18 What is the map distance between sn and ct?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, the dominant Bar mutation (B, chromosome X, position 57) results in thin bar- shaped eyes, while the recessive singed (sn, chromosome X, position 21) results burnt looking bristles. True breeding, wild type females are mated with true breeding males with Bar eyes and singed bristles. Using Drosophila notation, diagram the P1 and F1 crosses. P1 F1 Fill in the chart with phenotypic ratios that would be expected in the F2 generation. Use the space provided to show your work. Phenotype Females Males Overall (♀and ♂) =1 =1 =1arrow_forward
- In Drosophila, the allele for red eyes (pt) is wild-type and the allele for purple eyes (p¯) is mutant. The allele for grey body (b+) is wild-type and the allele for black body (b¯) is mutant. Flies with p*p* b*b* genotypes are mated with flies that have p p- b¯b¯ genotypes. A testcross was then performed in which the F1 offspring with p*p¯ b*b¯ genotypes were mated with flies with p p¯ b¯b genotypes. Ten-thousand flies were produced from this test cross. The following results were observed: 4,300 red eye, grey body flies 550 red eye, black body flies 4,500 purple eye, black body flies 650 purple eye, grey body flies Which F2 phenotypes are parental types? Which F2 phenotypes are recombinant types? What is the distance between the gene loci for eye color and body color? Use the equation for Hardy-Weinberg equilibrium for the following questions. p+q = 1 p2 + 2pq + q2 = 1arrow_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 wild type male. The resulting progeny were allowed to mate with one another to produce the data set. Three repetitions of the experiment were conducted. The following data were produced from the crosses. Test these data to determine if they are significantly different from the expected phenotypic ratio. Use the 5% level of significance. Your answer should include the hypothesized cross in genotypes, the Chi-squared value, the critical value and whether you reject or do not reject for each experiment. Wild eye Wild body – 112, Wild eye Ebony body – 40, Sepia eye Wild body – 35, Sepia eye Ebony body – 11arrow_forwardIn Drosophila, a female with three recessive traits, miniature wings, (m), ebony body (e) and rough eyes (r) were crossed with true- breeding wild-type males. In the F₁, all offspring had wild-type phenotypes. F, offspring were testcrossed to miniature, ebony, rough individuals and 1000 offspring were counted, shown in the table below. Phenotype Offspring m e r 310 ++ + 284 + e r m+ + m + r + e + m e + + + r Total 150 156 45 32 8 15 1000 1.Determine the genotypes of the P, and F, parents. 2.Determine the sequence of the three genes and the map distance between them. 3.Are there more or fewer double crossovers than expected? 4.Calculate the interference.arrow_forward
- Female Drosophila with cinnabar eye (cn) and vestigal wings (vg) were mated to males with roof wings (fr). The F1 were all wild type. When the F1 females were tested and crossed with male homozygous for all thrre traits, the following result were obtained. For this problem, rf, cn, vg is for roof wing / cinnabar eye / vestugal wing rf + cn + vg all wild type (no definite phenotype described). What is the value of coincidence and interference?arrow_forwardIn Drosophila, the brown mutation (bw, chromosome 2, position 104.5) results in brown eyes, while miniature (min, chromosome X, position 36.1) results in wings that are 2/3 the length of wild type. True breeding, wild type females are mated with true breeding males with brown eyes and miniature wings. Using Drosophila notation, diagram the P1 and F1 crosses. P1 F1 Fill in the chart with phenotypic ratios that would be expected in the F2 generation. Use the space provided to show your work. Phenotype Females Males Overall (♀and ♂) =1 =1 =1arrow_forwardA cross in Drosophila melanogaster involved the recessive X-linked genes for white eye (w), yellow body (y), and cut wings (c). A wild-type tri-hybrid female was crossed with wild-type males and only the male offspring were tallied. On the basis of the results shown below, which of the choices shown best represents the genetic map of the three loci on the X-chromosome? Phenotype Male Offspring + у t 494 394 + + ct 28 w y + 35 + у + 105 w + ct 101 w y ct 5 5.4 mu 17.6 mu ct 6.1 mu 18.3 mu w ct 17.6 mu 5.4 mu ct 18.3 mu 6.1 mu ct 6.1 mu 18.3 mu ct Submit Request Answer de Feedback Next >arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Biology (MindTap Course List)BiologyISBN:9781337392938Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. BergPublisher:Cengage Learning
Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning
Concepts of BiologyBiologyISBN:9781938168116Author:Samantha Fowler, Rebecca Roush, James WisePublisher:OpenStax College
Biology (MindTap Course List)
Biology
ISBN:9781337392938
Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg
Publisher:Cengage Learning
Human Heredity: Principles and Issues (MindTap Co...
Biology
ISBN:9781305251052
Author:Michael Cummings
Publisher:Cengage Learning
Concepts of Biology
Biology
ISBN:9781938168116
Author:Samantha Fowler, Rebecca Roush, James Wise
Publisher:OpenStax College
How to solve genetics probability problems; Author: Shomu's Biology;https://www.youtube.com/watch?v=R0yjfb1ooUs;License: Standard YouTube License, CC-BY
Beyond Mendelian Genetics: Complex Patterns of Inheritance; Author: Professor Dave Explains;https://www.youtube.com/watch?v=-EmvmBuK-B8;License: Standard YouTube License, CC-BY