In Drosophila, raspberry eyes (ras) is a sex-linked trait (X chromosome) and recessive. Polished eyes (pol) is autosomal and recessive. A raspberry, polished male is crossed with a homozygous wild-type female, producing an F1 in which both males and females are phenotypically wild-type. The F1 are inbred to form an F2. At what frequency (the number of total flies) would you expect to find a phenotypically raspberry (but not polished) male in the F2? 1. 1/16 2. 3/16 3. 1/4 4. 6/16 5.9/16

In Drosophila, raspberry eyes (ras) is a sex-linked trait (X chromosome) and recessive. Polished eyes (pol) is autosomal and recessive. A raspberry, polished male is crossed with a homozygous wild-type female, producing an F1 in which both males and females are phenotypically wild-type. The F1 are inbred to form an F2. At what frequency (the number of total flies) would you expect to find a phenotypically raspberry (but not polished) male in the F2? 1. 1/16 2. 3/16 3. 1/4 4. 6/16 5.9/16

Name: In Drosophila, raspberry eyes (ras) is a sex-linked trait (X chromosom
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Description: In Drosophila, raspberry eyes (ras) is a sex-linked trait (X chromosome) and recessive. Polished eyes (pol) is autosomal and recessive. A raspberry, polished male is crossed with a homozygous wild-type female, producing an F1 in which both males and

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter15: Genomes And Genomics

Section: Chapter Questions

Problem 2QP: Hemophilia and color blindness are both recessive conditions caused by genes on the X chromosome. To...

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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?
In Drosophila, ebony body colour is produced by a recessive gene a and wild-type (gray) body colour by its dominant allele a+. Vestigial wings are governed by a recessive gene vg, and normal wing size (wild type) by its dominant allele vg+. If wild-type dihybrid flies are crossed and produce 256 progeny, how many of these progeny flies are expected in each phenotypic class?
In Drosophila, raspberry eyes (ras) is a sex-linked and recessive trait. Polished eyes (pol) is autosomal and recessive. A raspberry, polished male is crossed with a homozygous wild-type female, producing an F1 in which both males and females are phenotypically wild-type. The F1 are inbred to form an F2. At what frequency would you expect to find a phenotypically raspberry polished female in the F2? 1. 1/16 2. 3/16 3. 6/16 4. 9/16 5. never
The recessive allele s causes Drosophila to have small wings, and the s+ allele causes normal wings. This gene is known to be X linked. If a small-winged male is crossed with a homozygous wild-type female, what ratio of normal to small-winged flies can be expected in each sex in the F1? If F1 flies are intercrossed, what F2 progeny ratios are expected? What progeny ratios are predicted if F1 females are backcrossed with their father?
The genes for the recessive traits of mahogany eyes and ebony body are approximately 30 map units apart on chromosome III in Drosophila. Assume that a mahogany-eyed female was mated to an ebony-bodied male and that the resulting F1 females (all phenotypically wild-type) were then mated to mahogany, ebony males. Of 1000 offspring, what would be the expected phenotypes, and in what numbers would they be expected? Group of answer choices None of these is correct wild-type = 200; mahogany and ebony = 200; mahogany = 300; ebony = 300 mahogany = 200; ebony = 200; wild-type = 300; mahogany and ebony = 300 wild-type = 350; mahogany and ebony = 350; mahogany = 150; ebony = 150 mahogany = 350; ebony = 350; wild-type = 150; mahogany and ebony = 150
The phenotype of crooked wings (cw) in Drosophila melanogaster is caused by a recessive mutant gene that independently assorts with a recessive mutant gene for hairy (h) body. Assume that a cross is made between a fly with normal wings and a hairy body and a fly with crooked wings and normal body hair. All F1 flies from this cross were wild-type, and these flies were crossed among each other to produce 288 F2 offspring. Which phenotypes would you expect among the offspring in the F2 generation, and how many of each phenotype would you expect?
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?
In Drosophila, males from a true-breeding stock with raspberry-colored eyes (instead of normal brownish-red eyes) were mated to females from a true-breeding stock with sable- colored bodies (instead of normal brown bodies). In the F, 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 F, males and females were mated, the F generation was composed of: 216 females with wild-type eyes and 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. Explain these results by diagramming the crosses and calculating any relevant map distances.
The 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 hairy
You have a Drosophila line that is homozygous for autosomal recessive alleles a, b, and c, linked in that order. Youcross females of this line with males homozygous for thecorresponding wild-type alleles. You then cross the F1 heterozygous males with their heterozygous sisters. You obtain the following F2 phenotypes (where letters denoterecessive phenotypes and pluses denote wild-type phenotypes): 1364 + + +, 365 a b c, 87 a b +, 84 + + c,47 a + +, 44 + b c, 5 a + c, and 4 + b +.a. What is the recombinant frequency between a andb? Between b and c? (Remember, there is no crossingover in Drosophila males.)b. What is the coefficient of coincidence?
In Drosophila, white eyes (w) are recessive to red eyes (w+) at one locus and black body (b) is recessive to gray body (b+). A homozygous white eyes, gray bodied female is crossed with a homozygous red eyes, black bodied male to produce the F1 progeny. The F1 progeny are testcrossed and produce the following progeny: White eyes, black body: 212 White eyes, gray body: 288 Red eyes, black body: 308 Red eyes, gray body: 192 Does the evidence indicate that w and b loci are linked? Explain why or why not? If they are linked, what is the map distance between the two loci? If they are not linked, what is the map distance between the two loci? If they are linked, are the allels in the F1 in coupling or repulsion? How do you know? Draw the genotypes of all individuals described in the problem (original parents, F1, testcross, and F2 progeny) using the appropriate notation.
In Drosophila, vermilion eye color is due to a recessive allele (v) located on the X chromosome. Curved wings are due to a recessive allele (cu) located on one autosome, and ebony body is due to a recessive allele (e) located on another autosome. A vermilion male is mated to a curved, ebony female, and the F1 males are phenotypically wild-type. If these males were backcrossed to curved, ebony females, what proportion of the F2 offspring will be wild-type males?